Chapter 6 : Co-ordinate Geometry - Rd Sharma Solutions for Class 10 Maths CBSE

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Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.1

Question 1

Solution 1

Question 2

Solution 2

  

Question 3

Solution 3



Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9



Question 10

Show that the points A(1,-2), B(3,6), C(5,10) and D(3,2) are the vertices of a parallelogram.

Solution 10

Question 11

Prove that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square.

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of triangle ABC right angled at B. Find the values of a and hence the area of ABC.

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21



Question 22

Solution 22

Question 23

Solution 23



Question 24

Find a point which is equidistant from the points

A (-5, 4) and B (-1, 6). How many such points are there?

Solution 24

Question 25

The centre of a circle is (2a, a - 7). Find the values of a if the circle passes through the point (11, -9) and has diameter  units.

Solution 25

Question 26

Ayush starts walking from his house to office, Instead of going to the office directly, he goes to bank first, from there to his daughter's school and then reaches the office. What is the extra distance travelled by Ayush in reaching the office? (Assume that all distance covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8) school at (13, 14) and office at (13, 26) and coordinates are in kilometer.

Solution 26

Question 27

Solution 27

Question 28

If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the (i) interior (ii) exterior of the triangle.

Solution 28

 

Question 29

Solution 29

Question 30

Solution 30



Question 31

Solution 31

Question 32

Show that the points A(5,6), B(1,5), C(2,1) and D(6,2) are the vertices of a square.

Solution 32





Question 33

Prove that the points A(2, 3), B(-2, 2), C(-1, -2), and D (3, -1) are the vertices of a square ABCD.

Solution 33



table attributes columnalign left end attributes row cell A B equals square root of left parenthesis 2 minus left parenthesis negative 2 right parenthesis right parenthesis squared plus left parenthesis 3 minus 2 right parenthesis squared end root equals square root of 4 squared plus 1 squared end root end cell row cell B C equals square root of left parenthesis negative 2 minus left parenthesis negative 1 right parenthesis right parenthesis squared plus left parenthesis 2 minus left parenthesis negative 2 right parenthesis right parenthesis squared end root equals square root of 1 squared plus 4 squared end root end cell row cell C D equals square root of left parenthesis negative 1 minus 3 right parenthesis squared plus left parenthesis negative 2 minus left parenthesis negative 1 right parenthesis right parenthesis squared end root equals square root of 4 squared plus 1 squared end root end cell row cell A D equals square root of left parenthesis 2 minus 3 right parenthesis squared plus left parenthesis 3 minus left parenthesis negative 1 right parenthesis right parenthesis squared end root equals square root of 1 squared plus 4 squared end root end cell row cell A B equals B C equals C D equals A D end cell row cell text Now   let   us   find   the   lengths   of   the   diagonals   AC   and   BD. end text end cell row cell text AC = end text square root of left parenthesis 2 minus left parenthesis negative 1 right parenthesis right parenthesis squared plus left parenthesis 3 minus left parenthesis negative 2 right parenthesis right parenthesis squared end root equals square root of 3 squared plus 5 squared end root equals square root of 34 end cell row cell B D equals square root of left parenthesis left parenthesis negative 2 right parenthesis minus 3 right parenthesis squared plus left parenthesis 2 minus left parenthesis negative 1 right parenthesis right parenthesis squared end root equals square root of 5 squared plus 3 squared end root equals square root of 34 end cell row cell text For   ABCD ,  all   sides   are   of   equal   length   and end text end cell row cell text the   lengths   of   the   diagonals   are   also   equal. end text end cell row cell text Hence ,  ABCD   is   a   square. end text end cell end table

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

If the point P(x,y) is equidistant from the points A(5,1) and B(1,5), prove that x = y.

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

If the point P(k-1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.

Solution 41

table attributes columnalign left end attributes row cell text We   know   that end text end cell row cell text PA  =  end text square root of left parenthesis k minus 1 minus 3 right parenthesis squared plus left parenthesis 2 minus k right parenthesis squared end root end cell row cell text and end text end cell row cell text PB  =  end text square root of left parenthesis k minus 1 minus k right parenthesis squared plus left parenthesis 2 minus 5 right parenthesis squared end root end cell row cell text Given   that   PA  =  PB end text end cell row cell text Squaring   both   the   sides , end text end cell row cell P A squared equals P B squared end cell row cell rightwards double arrow left parenthesis k minus 1 minus 3 right parenthesis squared plus left parenthesis 2 minus k right parenthesis squared equals left parenthesis k minus 1 minus k right parenthesis squared plus left parenthesis 2 minus 5 right parenthesis squared end cell row cell rightwards double arrow left parenthesis k minus 4 right parenthesis squared plus left parenthesis 2 minus k right parenthesis squared equals left parenthesis negative 1 right parenthesis squared plus left parenthesis 3 right parenthesis squared end cell row cell rightwards double arrow k squared plus 16 minus 8 k plus 4 plus k squared minus 4 k equals 1 plus 9 end cell row cell rightwards double arrow 2 k squared minus 12 k plus 10 equals 0 end cell row cell rightwards double arrow k squared minus 6 k plus 5 equals 0 end cell row cell rightwards double arrow k squared minus 5 k minus k plus 5 equals 0 end cell row cell rightwards double arrow k left parenthesis k minus 5 right parenthesis minus 1 left parenthesis k minus 5 right parenthesis equals 0 end cell row cell rightwards double arrow left parenthesis k minus 5 right parenthesis left parenthesis k minus 1 right parenthesis equals 0 end cell row cell rightwards double arrow k equals 5 text   or   end text k equals 1 end cell end table

Question 42

If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also, find the length of AB.

Solution 42

table attributes columnalign left end attributes row cell text We   know   that , end text end cell row cell text AB  =  end text square root of left parenthesis 0 minus 3 right parenthesis squared plus left parenthesis 2 minus p right parenthesis squared end root... left parenthesis 1 right parenthesis end cell row cell text and end text end cell row cell text AC  =  end text square root of left parenthesis 0 minus p right parenthesis squared plus left parenthesis 2 minus 5 right parenthesis squared end root end cell row cell text Given   that   AB  =  AC end text end cell row cell text Squaring   both   the   sides ,  we   have , end text end cell row cell A B squared equals A C squared end cell row cell rightwards double arrow left parenthesis 0 minus 3 right parenthesis squared plus left parenthesis 2 minus p right parenthesis squared equals left parenthesis 0 minus p right parenthesis squared plus left parenthesis 2 minus 5 right parenthesis squared end cell row cell rightwards double arrow 9 plus 4 plus p squared minus 4 p equals p squared plus 9 end cell row cell rightwards double arrow 4 p equals 4 end cell row cell rightwards double arrow p equals 1 end cell row cell text Substituting   the   value   p  =  1   in   equation  ( 1 ), end text end cell row cell A B equals square root of left parenthesis 0 minus 3 right parenthesis squared plus left parenthesis 2 minus p right parenthesis squared end root end cell row cell equals square root of 9 plus 1 end root equals square root of 10 end cell end table

Question 43

Solution 43



Question 44

Find the equation of the perpendicular bisector of the line segment joining points (7,1) and (3,5).

Solution 44

Question 45

Prove that the points (3,0), (4,5), (-1,4) and (-2,-1), taken in order, form a rhombus. Also, find its area.

Solution 45



Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

If a point A(0,2) is equidistant from the points B(3,p) and C(p,5), then find the value of p.

Solution 49

Question 50

Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.

Solution 50

 


table attributes columnalign left end attributes row cell text Let   the   points   be   A ,  B   and   C   respectively. end text end cell row cell text AB = end text square root of left parenthesis text 7 end text minus left parenthesis negative 2 right parenthesis right parenthesis squared plus left parenthesis 10 minus 5 right parenthesis squared end root equals square root of 9 squared plus 5 squared end root equals square root of 106 end cell row cell B C equals square root of left parenthesis negative 2 minus 3 right parenthesis squared plus left parenthesis 5 minus left parenthesis negative 4 right parenthesis right parenthesis squared end root equals square root of 5 squared plus 9 squared end root equals square root of 106 end cell row cell C A equals square root of left parenthesis 3 minus 7 right parenthesis squared plus left parenthesis negative 4 minus 10 right parenthesis squared end root equals square root of 4 squared plus 14 squared end root equals square root of 212 equals square root of 2 cross times 106 end root end cell row cell rightwards double arrow A B squared plus B C squared equals C A squared end cell row cell text Since   AB = BC   and   end text A B squared plus B C squared equals C A squared comma text   end text capital delta text ABC   is   an end text end cell row cell text isosceles   right   triangle. end text end cell end table



Question 51

If the point P(x, 3) is equidistant from the points A(7, -1) and B(6, 8), find the value of x and the find the distance AP.

Solution 51



table attributes columnalign left end attributes row cell text We   know   that   end text end cell row cell text PA = end text square root of left parenthesis x minus 7 right parenthesis squared plus left parenthesis 3 minus left parenthesis negative 1 right parenthesis right parenthesis blank presuperscript 2 end root end cell row cell text and end text end cell row cell text PB = end text square root of left parenthesis x minus 6 right parenthesis squared plus left parenthesis 3 minus 8 right parenthesis squared end root end cell row cell text Given   that   PA  =  PB end text end cell row cell text Squaring   both   the   sides ,  we   have , end text end cell row cell P A squared equals P B squared end cell row cell rightwards double arrow left parenthesis x minus 7 right parenthesis squared plus left parenthesis 3 minus left parenthesis negative 1 right parenthesis right parenthesis equals presuperscript 2 left parenthesis x minus 6 right parenthesis squared plus left parenthesis 3 minus 8 right parenthesis squared end cell row cell rightwards double arrow x squared plus 49 minus 14 x plus 16 equals x squared plus 36 minus 12 x plus 25 end cell row cell rightwards double arrow 2 x equals 4 end cell row cell rightwards double arrow x equals 2 end cell row cell text Substituting   the   value   x  =  2   in   PA ,  we   get , end text end cell row cell P A equals square root of left parenthesis 2 minus 7 right parenthesis squared plus left parenthesis 3 minus left parenthesis negative 1 right parenthesis right parenthesis blank presuperscript 2 end root equals square root of 5 squared plus 4 squared end root equals square root of 41 end cell end table

Question 52

If A(3, y) is equidistant from the points P(8, - 3) and Q(7, 6), find the value of y and find the distance AQ.

Solution 52



table attributes columnalign left end attributes row cell text We   know   that   end text end cell row cell text AP = end text square root of left parenthesis 3 minus 8 right parenthesis squared plus left parenthesis y minus left parenthesis negative 3 right parenthesis right parenthesis blank presuperscript 2 end root end cell row cell text and end text end cell row cell text AQ = end text square root of left parenthesis 3 minus 7 right parenthesis squared plus left parenthesis y minus 6 right parenthesis squared end root... left parenthesis 1 right parenthesis end cell row cell text Given   that   AP  =  AQ end text end cell row cell text Squaring   both   the   sides ,  we   get , end text end cell row cell A P squared equals A Q squared end cell row cell rightwards double arrow left parenthesis 3 minus 8 right parenthesis squared plus left parenthesis y minus left parenthesis negative 3 right parenthesis right parenthesis equals presuperscript 2 left parenthesis 3 minus 7 right parenthesis squared plus left parenthesis y minus 6 right parenthesis squared end cell row cell rightwards double arrow 25 plus y squared plus 9 plus 6 y equals 16 plus y squared plus 36 minus 12 y end cell row cell rightwards double arrow 18 y equals 18 end cell row cell rightwards double arrow y equals 1 end cell row cell text Substituting   the   value   y  =  1   in   equation  ( 1 ),  we   get end text end cell row cell A Q equals square root of left parenthesis 3 minus 7 right parenthesis squared plus left parenthesis 1 minus 6 right parenthesis squared end root equals square root of 4 squared plus 5 squared end root equals square root of 41 end cell end table

Question 53

If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex.

Solution 53

For an equilateral triangle, the perpendicular bisector of any side passes through the opposite vertex.

Both the points, (0, -3) and (0, 3), lie on the y-axis equidistant from the origin. Hence, the perpendicular bisector joining these two points is the x-axis.

Any point on the x-axis has the coordinates (a, 0).

The distance between (0, -3) and (0, 3) is 6.

Hence, the distance between (a, 0) and (0, 3) should also be 6.

62 = (a - 0)2 + (0 - 3)2

36 = a2 + 9

a2 = 27


a equals plus-or-minus 3 square root of 3

text Hence ,  the   coordinates   of   the   third   vertex   are   end text left parenthesis plus-or-minus text 3 end text square root of text 3 end text end root comma 0 right parenthesis.


Question 54

If the point P(2, 2) is equidistant from the points A(-2, k) and B(-2k, -3), find k. Also, find the length of AP.

Solution 54

table attributes columnalign left end attributes row cell text We   know   that , end text end cell row cell text PA  =  end text square root of left parenthesis 2 minus left parenthesis negative 2 right parenthesis right parenthesis squared plus left parenthesis 2 minus k right parenthesis squared end root end cell row cell text and end text end cell row cell text PB  =  end text square root of left parenthesis 2 minus left parenthesis negative 2 k right parenthesis right parenthesis squared plus left parenthesis 2 minus left parenthesis negative 3 right parenthesis right parenthesis squared end root end cell row cell text Given   that   PA  =  PB end text end cell row cell text Squaring   both   the   sides ,  we   get , end text end cell row cell P A squared equals P B squared end cell row cell rightwards double arrow left parenthesis 2 minus left parenthesis negative 2 right parenthesis right parenthesis squared plus left parenthesis 2 minus k right parenthesis squared equals left parenthesis 2 minus left parenthesis negative 2 k right parenthesis right parenthesis squared plus left parenthesis 2 minus left parenthesis negative 3 right parenthesis right parenthesis squared end cell row cell rightwards double arrow 4 squared plus 4 plus k squared minus 4 k equals 4 plus 4 k squared plus 8 k plus 25 end cell row cell rightwards double arrow 3 k squared plus 12 k plus 9 equals 0 end cell row cell rightwards double arrow k squared plus 4 k plus 3 equals 0 end cell row cell rightwards double arrow k squared plus 3 k plus k plus 3 equals 0 end cell row cell rightwards double arrow k left parenthesis k plus 3 right parenthesis plus 1 left parenthesis k plus 3 right parenthesis equals 0 end cell row cell rightwards double arrow left parenthesis k plus 3 right parenthesis left parenthesis k plus 1 right parenthesis equals 0 end cell row cell rightwards double arrow k equals negative 3 text   or   end text k equals negative 1 end cell row cell F o r text   end text k equals negative 3 comma text   A end text identical to left parenthesis negative 2 comma negative 3 right parenthesis end cell row cell A P equals square root of left parenthesis 2 minus left parenthesis negative 2 right parenthesis right parenthesis squared plus left parenthesis 2 minus left parenthesis negative 3 right parenthesis right parenthesis squared end root equals square root of 4 squared plus 5 squared end root equals square root of 41 end cell row cell F o r text   end text k equals negative 1 comma text   A end text identical to left parenthesis negative 2 comma negative 1 right parenthesis end cell row cell A P equals square root of left parenthesis 2 minus left parenthesis negative 2 right parenthesis right parenthesis squared plus left parenthesis 2 minus left parenthesis negative 1 right parenthesis right parenthesis squared end root equals square root of 4 squared plus 3 squared end root equals square root of 25 equals 5 end cell end table

Question 55

Show that ∆ABC, where A (-2, 0), B (2, 0) C (0, 2) and ∆PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar.

Solution 55

Using the distance formula,

AB=

AC=

BC=

PQ=

PR=

QR=

Now,

ΔABC ~ ΔPQR

by the SSS test.

Question 56

Solution 56


Question 57

Find the circumcentre of the triangle whose vertices are (-2,-3), (-1,0), (7,-6).

Solution 57

Question 58

Solution 58

Question 59

Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).

Solution 59

Question 60

Solution 60



Question 61

Solution 61



Question 62

Solution 62

Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.3

Question 1

Solution 1

Question 2

Find the points of trisection of the line segment joining the points:

(i) (5, -6) and (-7, 5), (ii) (3, -2) and (-3,-4), (iii) (2,-2) and (-7,4).

Solution 2

(i)



(ii)



(iii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

If P(9a - 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.

Solution 5

Question 6

If (a, b) is the mid-point of the line segment joining the points A (10, -6), B(k, 4) and a - 2b = 18, find the value of k and the distance AB.

Solution 6

Question 7

Solution 7

Question 8

Solution 8



Question 9

If the points P,Q(x, 7), R, S(6, y) in this order divide the line segment joining A(2, p) and B (7, 10) in 5 equal parts, find x, y and p.

Solution 9

 

  

 

 

  

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

  

Solution 14

  

Question 15

Solution 15

 
 
*Note: (i) Answer given in the book is incorrect.
Question 16

Solution 16

Question 17

Prove that (4,3), (6,4), (5,6) and (3,5) are the angular points of a square.

Solution 17



Question 18

Solution 18



Question 19

Solution 19

Question 20

Find the ratio in which the line segment joining the points A (3, -3) and B (-2, 7) is divided by x- axis. Also, find the coordinates of the point of division.

Solution 20

Let the point on the x-axis be (a, 0).

Let this point divide the line segment AB in the ratio of r : 1.

Using the section formula for the y-coordinate, we get

 



table attributes columnalign left end attributes row cell 0 equals fraction numerator 7 r minus 3 over denominator r plus 1 end fraction end cell row cell rightwards double arrow 7 r minus 3 equals 0 end cell row cell rightwards double arrow r equals 3 over 7 end cell row cell text Hence ,  the   line   segment   joining   A   and   B   end text end cell row cell text is   divided   in   the   ratio   of   3  :  7   by   the   x-axis. end text end cell end table

Question 21

Find the ratio in which the point P(x, 2) divides the line segment joining the points A (12, 5) and B (4, -3). Also, find the value of x.

Solution 21

 



table attributes columnalign left end attributes row cell text Let   P   divide   the   line   joining   A   and   B end text end cell row cell text   in   the   ratio   of   r  :  1 end text end cell row cell text Using   the   section   formula   for   the   y-coordinate ,  we   get end text end cell row cell text 2 = end text fraction numerator negative 3 text r + 5 end text over denominator text r + 1 end text end fraction end cell row cell rightwards double arrow 2 r plus 2 equals negative 3 r plus 5 end cell row cell rightwards double arrow 5 r equals 3 end cell row cell rightwards double arrow r equals 3 over 5 end cell row cell text Hence ,  P   divides   the   line   joining   A   and   B end text end cell row cell text   in   the   ratio   of   3  :  5 end text end cell row cell text Using   the   section   formula   for   the   x-coordinate ,  we   get end text end cell row cell x equals fraction numerator 12 plus 60 over denominator 8 end fraction equals 72 over 8 equals 9 end cell end table

Question 22

Find the ratio in which the point P(-1, y) lying on the line segment joining A(-3, 10) and B(6, -8) divides it. Also find the value of y.

Solution 22




table attributes columnalign left end attributes row cell text Let   P   divide   A   and   B   in   the   ratio   of   r  :  1 end text end cell row cell text P end text identical to left parenthesis negative 1 comma y right parenthesis comma text   A end text identical to left parenthesis negative 3 comma 10 right parenthesis comma text   end text B identical to left parenthesis 6 comma negative 8 right parenthesis end cell row cell text Using   the   section   formula   for   x   coordinate ,  we   get end text end cell row cell negative 1 equals fraction numerator 6 r minus 3 over denominator r plus 1 end fraction rightwards double arrow negative r minus 1 equals 6 r minus 3 end cell row cell 7 r equals 2 rightwards double arrow r equals 2 over 7 end cell row cell text Hence ,  P   divides   the   line   AB   in   the   ratio   of   2  :  7 end text end cell row cell text Hence ,  using   the   section   formula , end text end cell row cell text y = end text fraction numerator negative text 8 r + 10 end text over denominator r plus 1 end fraction end cell row cell rightwards double arrow therefore y equals fraction numerator negative 16 plus 70 over denominator 2 plus 7 end fraction equals 54 over 9 equals 6 text        end text left square bracket text Substituting   r end text equals 2 over 7 right square bracket end cell end table

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

In what ratio does the point (-4,6) divide the line segment joining the points A(-6,10) and B(3,-8)?

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29



Question 30

Solution 30



Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34



Question 35

Show that the points A(1,0) B(5,3), C(2,7) and D(-2,4) are the vertices of a parallelogram.

Solution 35



Question 36

Solution 36



Question 37

Solution 37



Question 38

Solution 38



Question 39

Points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in 5 equal parts. Find the coordinates of the points P, Q, and R.

Solution 39

The difference between the x-coordinates of A and B is 6 - 1 = 5

Similarly, the difference between the y-coordinates of A and B is 7 - 2 = 5

Hence, if the line segment joining A(1, 2) and B(6, 7) is divided into 5 equal parts by the points P, Q, R and S, then the coordinates of P, Q, R and S can be found out by increasing the x and the y coordinates of A by 1 successively.

Hence, the coordinates of P are (1 + 1, 2 + 1) = (2, 3)

The coordinates of Q are (2 + 1, 3 + 1) = (3, 4)

The coordinates of R are (3 + 1, 4 + 1) = (4, 5)

 

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

If two vertices of a parallelogram are (3,2), (-1,0) and the diagonals cut at (2,-5), find the other vertices of the parallelogram.

Solution 47



Question 48

If the coordinates of the mid-points of the sides of a triangle are (3,4), (4,6), and (5,7), find its vertices.

Solution 48



Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52





Question 53

Solution 53



Question 54

Solution 54

Question 55

Solution 55






Question 56

Solution 56

Question 57

A point P divides the line segment joining the points A(3,-5) and B(-4,8), such that  . If P lies on the line x + y = 0, then find the value of k.

Solution 57

Given points are A(3,-5) and B(-4,8).

P divides AB in the ratio k : 1.


Using the section formula, we have:

Coordinate of point P are {(-4k+3/k+1)(8k-5/k+1)}

Now it is given, that P lies on the line x+y = 0

Therefore,

-4k+3/k+1 + 8k-5/k+1 =0

=> -4k+3+8k-5 =0

=> 4k -2 =0

=> k=2/4

=> k=1/2

Thus, the value of k is 1/2.

Question 58

The mid - point P of the line segment joining the points A (-10, 4) and B (-2, 0) lies on the line segment joining the points C (-9, -4) and D (-4, y). Find the ratio in which P divides CD. Also, find the value of y.

Solution 58




table attributes columnalign left end attributes row cell text Coordinates   of   the   midpoint   P   of   A   and   B   are end text end cell row cell left parenthesis fraction numerator negative 10 plus left parenthesis negative 2 right parenthesis over denominator 2 end fraction comma text   end text fraction numerator 4 plus 0 over denominator 2 end fraction right parenthesis identical to left parenthesis negative 6 comma text   2 end text right parenthesis end cell row cell text P   lies   on   the   line   joining   C   and   D. end text end cell row cell text Let   P end text left parenthesis negative 6 comma 2 right parenthesis text   divide   C end text left parenthesis negative 9 comma negative 4 right parenthesis text   and   D end text left parenthesis negative 4 comma y right parenthesis end cell row cell text   in   the   ratio   of   r  :  1 end text end cell row cell text Using   the   section   formula   for   the   x-coordinate   we   get end text end cell row cell negative 6 equals fraction numerator negative 4 r minus 9 over denominator r plus 1 end fraction rightwards double arrow negative 6 r minus 6 equals negative 4 r minus 9 end cell row cell rightwards double arrow 2 r equals 3 rightwards double arrow r equals 3 over 2 end cell row cell text Hence ,  P end text left parenthesis negative 6 comma 2 right parenthesis text   divides   C end text left parenthesis negative 9 comma negative 4 right parenthesis text   and   D end text left parenthesis negative 4 comma y right parenthesis end cell row cell text in   the   ratio   of   3  :  2 end text end cell row cell text Using   the   section   formula   for   y-coordinate   we   get end text end cell row cell 2 equals fraction numerator 3 y minus 8 over denominator 3 plus 2 end fraction rightwards double arrow 10 equals 3 y minus 8 rightwards double arrow 3 y equals 18 end cell row cell rightwards double arrow y equals 6 end cell end table

Question 59

If the point C(-1, 2) divides internally the line segment joining the points A(2, 5) and B(x, y) in the ratio 3 : 4, find the value of x2 + y2.

Solution 59

Question 60

ABCD is a parallelogram with vertices A(x1, y1), B(x2, y2) and C (x3, y3). Find the coordinates of the fourth vertex D in terms of x1, x2, x3, y1, y2 and y3.

Solution 60

Question 61

The points A (x1, y1), B(x2, y2) and C (x3, y3) are the vertices of ABC.

 i. The median from A meets BC at D. Find the coordinates of the point D.

 ii. Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

 iii. Find the points of coordinates Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.

 iv. What are the coordinates of the centroid of the triangle ABC

Solution 61

Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.

Solution 7



Question 8

Solution 8



Question 9

Solution 9


Question 10

Solution 10

Question 11



Solution 11



Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3



Question 4

Find the area of the quadrilaterals, the coordinates of whose vertices are

(-4,-2), (-3,-5), (3,-2), (2,3)

Solution 4




Question 5

Solution 5

Question 6

Solution 6



Question 7

Solution 7

Question 8

Find the area of a quadrilateral ABCD, the coordinates of whose vertices are A(-3, 2), B(5, 4), C(7, -6) and D(-5, -4).

Solution 8

 

Question 9

In ABC, the coordinates of vertex A are (0, -1) and D(1, 0) and E(0, 1) respectively the mid-points of the sides AB and AC. If F is the mid-point of side BC, find the area of DEF.

Solution 9

  

Question 10

Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2).

Solution 10

 

  

 

 

Question 11

If P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.

Solution 11

begin mathsize 14px style Join space straight P space and space straight R.
Area space of space increment PSR equals 1 half open vertical bar negative 5 left parenthesis 2 plus 3 right parenthesis plus 1 left parenthesis negative 3 plus 3 right parenthesis plus 2 left parenthesis negative 3 minus 2 right parenthesis close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half open vertical bar negative 5 cross times 5 plus 1 cross times 0 plus 2 cross times left parenthesis negative 5 right parenthesis close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half open vertical bar negative 25 plus 0 minus 10 close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half open vertical bar negative 35 close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space equals 35 over 2
Area space of space increment PQR equals 1 half open vertical bar negative 5 left parenthesis negative 6 plus 3 right parenthesis minus 4 left parenthesis negative 3 plus 3 right parenthesis plus 2 left parenthesis negative 3 plus 6 right parenthesis close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half open vertical bar negative 5 cross times left parenthesis negative 3 right parenthesis minus 4 cross times 0 plus 2 cross times 3 close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half open vertical bar 15 plus 0 plus 6 close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space equals 21 over 2
Now comma space area space of space quadrilateral space PQRS equals Area space of space increment PSR plus Area space of space increment PQR
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 35 over 2 plus 21 over 2
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 35 plus 21 over denominator 2 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 56 over 2
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 28 space space sq. space units end style

Question 12

If A (-3, 5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.

Solution 12


Question 13

Solution 13

Question 14

Solution 14

Question 15

If the vertices of a triangle are (1,-3), (4,p) and (-9, 7) and its area is 15 sq. units, find the value (s) of p.

Solution 15



Let space straight A left parenthesis 1 comma negative 3 right parenthesis comma space straight B left parenthesis 4 comma straight p right parenthesis space and space straight C left parenthesis negative 9 comma 7 right parenthesis space be space the space vertices space of space increment ABC.
Area space of space increment ABC equals 15 space sq. space units
therefore 15 equals 1 half open vertical bar 1 left parenthesis straight p minus 7 right parenthesis plus 4 left parenthesis 7 plus 3 right parenthesis minus 9 left parenthesis negative 3 minus straight p right parenthesis close vertical bar
rightwards double arrow 15 equals 1 half open vertical bar straight p minus 7 plus 40 plus 27 plus 9 straight p close vertical bar
rightwards double arrow 15 equals 1 half open vertical bar 10 straight p plus 60 close vertical bar
rightwards double arrow 30 equals 10 straight p plus 60 space or space 30 equals negative 10 straight p minus 60
rightwards double arrow 10 straight p equals negative 30 space or space 10 straight p equals negative 90
rightwards double arrow straight p equals negative 3 space or space straight p equals negative 9

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(-2, 6) and C (3, 1) is 10 sequare units.

Solution 22

Question 23

If a ≠ b ≠ 0, prove that the points (a, a2), (b, b2), (0, 0) are never collinear.

Solution 23

Let the points (a, a2), (b, b2), (0, 0) represent a triangle. If we can prove that the area of the triangle so formed is not equal to zero, then we can prove that the points (a, a2), (b, b2), (0, 0) are never collinear.

Area of a triangle is given by

Now b≠a≠0.

So,

b - a≠0,

Δ≠0

Thus, points (a, a2), (b, b2), (0, 0) are never collinear.

Question 24

The area of a triangle is 5 sq. units. Two of its vertices are at (2, 1) and (3, -2). If the third vertex is (7/2, y), find y.

Solution 24

Area of a triangle is given by  

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27



Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31


table attributes columnalign left end attributes row cell F i n d text   end text t h e text   end text a r e a text   end text o f text   end text a text   end text p a r a l l e l o g r a m text   end text A B C D text   end text i f text   end text t h r e e text   end text o f text   end text end cell row cell i t s text   end text v e r t i c e s text   end text a r e text    A end text left parenthesis text 2 ,  4 end text right parenthesis text ,  B end text left parenthesis 2 plus square root of 3 comma 5 right parenthesis text    and    C end text left parenthesis text 2 ,  6 end text right parenthesis end cell end table

Solution 31


table attributes columnalign left end attributes row cell text Area   of   ABC   can   be   found   using   formula   for   area. end text end cell row cell capital delta text   =  end text fraction numerator text 1 end text over denominator text 2 end text end fraction vertical line x subscript 1 left parenthesis y subscript 2 minus y subscript 3 right parenthesis plus x subscript 2 left parenthesis y subscript 3 minus y subscript 1 right parenthesis plus x subscript 3 left parenthesis y subscript 1 minus y subscript 2 right parenthesis vertical line end cell row cell equals 1 half vertical line 2 left parenthesis 5 minus 6 right parenthesis plus left parenthesis 2 plus square root of 3 right parenthesis left parenthesis 6 minus 4 right parenthesis plus 2 left parenthesis 4 minus 5 right parenthesis vertical line end cell row cell equals 1 half vertical line minus 2 plus 4 plus 2 square root of 3 minus 2 vertical line end cell row cell equals square root of 3 text   square   units end text end cell row cell text The   triangle   formed   by   any   three   vertices   of   end text end cell row cell text a   parallelogram   has   half   the   area   of   the   parallelogram. end text end cell row cell text Hence ,  area   of   the   parallelogram = 2 end text cross times capital delta text = 2 end text square root of text 3 end text end root text   square   units end text end cell end table

Question 32

Find the value (s) of k for which the points (3k - 1, k - 2), (k, k - 7) and (k - 1, -k - 2) are collinear.

Solution 32


Let the points be A, B and C respectively.

If A, B and C are collinear, then the area of ∆ABC is zero.



table attributes columnalign left end attributes row cell capital delta text = end text fraction numerator text 1 end text over denominator text 2 end text end fraction vertical line x subscript 1 left parenthesis y subscript 2 minus y subscript 3 right parenthesis plus x subscript 2 left parenthesis y subscript 3 minus y subscript 1 right parenthesis plus x subscript 3 left parenthesis y subscript 1 minus y subscript 2 right parenthesis vertical line end cell row cell equals 1 half vertical line left parenthesis 3 k minus 1 right parenthesis left parenthesis k minus 7 plus k plus 2 right parenthesis plus k left parenthesis negative k minus 2 minus k plus 2 right parenthesis plus left parenthesis k minus 1 right parenthesis left parenthesis k minus 2 minus k plus 7 right parenthesis vertical line end cell row cell equals 1 half vertical line 4 k squared minus 12 k vertical line end cell end table




table attributes columnalign left end attributes row cell text The   area   of   the   triangle   is   zero   when   the   points end text end cell row cell text are   collinear. end text end cell row cell text 4 k end text to the power of text 2 end text end exponent minus 12 k equals 0 end cell row cell rightwards double arrow 4 k left parenthesis k minus 3 right parenthesis equals 0 end cell row cell rightwards double arrow k equals 0 text   or   end text k minus 3 equals 0 end cell row cell rightwards double arrow k equals 0 text   or   end text k equals 3 end cell end table

Question 33

If the points A(-1, -4), B(b, c) and C(5, -1) are collinear and 2b + c = 4, find the values of b and c.

Solution 33



table attributes columnalign left end attributes row cell text Let   the   points   be   A ,  B   and   C   respectively. end text end cell row cell text If   A ,  B   and   C   are   collinear ,  then   area   of   the   end text capital delta text ABC   is   zero. end text end cell row cell capital delta text   =  end text fraction numerator text 1 end text over denominator text 2 end text end fraction vertical line x subscript 1 left parenthesis y subscript 2 minus y subscript 3 right parenthesis plus x subscript 2 left parenthesis y subscript 3 minus y subscript 1 right parenthesis plus x subscript 3 left parenthesis y subscript 1 minus y subscript 2 right parenthesis vertical line end cell row cell equals 1 half vertical line left parenthesis negative 1 right parenthesis left parenthesis c minus left parenthesis negative 1 right parenthesis right parenthesis plus b left parenthesis negative 1 minus left parenthesis negative 4 right parenthesis right parenthesis plus 5 left parenthesis negative 4 minus c right parenthesis vertical line end cell row cell equals 1 half vertical line minus c minus 1 plus 3 b minus 20 minus 5 c vertical line end cell row cell equals 1 half vertical line 3 b minus 6 c minus 21 vertical line end cell row cell text The   area   of   the   triangle   is   zero   when   the   points end text end cell row cell text are   collinear. end text end cell row cell 3 b minus 6 c minus 21 equals 0 end cell row cell rightwards double arrow 3 left parenthesis negative b plus 2 c plus 7 right parenthesis equals 0 end cell row cell rightwards double arrow negative b plus 2 c plus 7 equals 0... left parenthesis 1 right parenthesis end cell row cell text Also   given   that   end text 2 b plus c equals 4... left parenthesis 2 right parenthesis end cell row cell text Solving   equations  ( 1 )  and  ( 2 ),  we   have , end text end cell row cell b equals 3 text   and   end text c equals negative 2 end cell end table

Question 34

If the points A(-2, 1), B(a, b) and C(4, -1) are collinear and a - b = 1, find the values of a and b.

Solution 34


table attributes columnalign left end attributes row cell text Let   the   points   be   A ,  B   and   C   respectively. end text end cell row cell text If   A ,  B   and   C   are   collinear ,  then   the   area   of   end text capital delta text ABC   is   zero. end text end cell row cell capital delta text = end text fraction numerator text 1 end text over denominator text 2 end text end fraction vertical line x subscript 1 left parenthesis y subscript 2 minus y subscript 3 right parenthesis plus x subscript 2 left parenthesis y subscript 3 minus y subscript 1 right parenthesis plus x subscript 3 left parenthesis y subscript 1 minus y subscript 2 right parenthesis vertical line end cell row cell equals 1 half vertical line left parenthesis negative 2 right parenthesis left parenthesis b minus left parenthesis negative 1 right parenthesis right parenthesis plus a left parenthesis negative 1 minus 1 right parenthesis plus 4 left parenthesis 1 minus b right parenthesis vertical line end cell row cell equals 1 half vertical line minus 2 b minus 2 minus 2 a plus 4 minus 4 b vertical line end cell row cell equals 1 half vertical line minus 2 a minus 6 b plus 2 vertical line end cell row cell text The   area   of   the   triangle   is   zero   when   the   points end text end cell row cell text are   collinear. end text end cell row cell negative 2 a minus 6 b plus 2 equals 0 end cell row cell rightwards double arrow negative 2 left parenthesis a plus 3 b minus 1 right parenthesis equals 0 end cell row cell rightwards double arrow a plus 3 b minus 1 equals 0... left parenthesis 1 right parenthesis end cell row cell text Also   given   that   end text a minus b equals 1... left parenthesis 2 right parenthesis end cell row cell text Solving   equations  ( 1 )  and  ( 2 ),  we   have , end text end cell row cell a equals 1 text   and   end text b equals 0 end cell end table

Question 35

If the points A(1, -2), B(2, 3), C(a, 2) and D(-4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

Solution 35

Question 36

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, find the area of ADE.

Solution 36

Question 37

If D(-1/5, 5/2), E(7, 3) and F(7/2, 7/2) are the mid-points of sides of ABC, find the area of ABC

Solution 37

  

 

Chapter 6 - Co-ordinate Geometry Exercise 6.63

Question 1

begin mathsize 12px style The space distance space between space the space points space left parenthesis cos space straight theta comma space sinθ right parenthesis space and space left parenthesis sin space straight theta space minus space cosθ right parenthesis space is
left parenthesis straight a right parenthesis space square root of 3
left parenthesis straight b right parenthesis space square root of 2
left parenthesis straight c right parenthesis space 2
left parenthesis straight d right parenthesis space 1 end style

Solution 1

begin mathsize 12px style space We space know comma space distance space between space two space point space left parenthesis straight x comma straight y right parenthesis space and space left parenthesis straight a comma straight b right parenthesis space is space calculated space as space
distance space equals space square root of left parenthesis straight x minus straight a right parenthesis squared plus left parenthesis straight y minus straight b right parenthesis squared end root
Here space points space are space left parenthesis cos space straight theta comma sin space straight theta right parenthesis space and space left parenthesis sin space straight theta space minus space cos space straight theta right parenthesis
Hence comma
distance space equals space square root of left parenthesis cosθ minus space sin space straight theta right parenthesis squared plus left parenthesis sinθ plus cosθ right parenthesis squared end root
space space space space space space space space space space space space space space space space space equals square root of left parenthesis cos squared straight theta space plus space sin squared straight theta plus 2 sinθcosθ plus sin squared straight theta plus cos squared straight theta minus 2 sinθcosθ right parenthesis end root space
We space know space sin squared straight theta plus cos squared straight theta space equals space 1
rightwards double arrow space distance space equals space square root of 1 plus 1 end root
space space space space space space space space space space space space space space space space space space space space space space space equals square root of 2 space space space space space space space
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 2

begin mathsize 12px style The space distance space between space the space points space left parenthesis straight a space cos 25 to the power of straight o comma 0 right parenthesis space and space left parenthesis 0 comma straight a space cos space 65 to the power of straight o right parenthesis space is space
left parenthesis straight a right parenthesis space straight a
left parenthesis straight b right parenthesis space 2 straight a
left parenthesis straight c right parenthesis space 3 straight a
left parenthesis straight d right parenthesis space None space of space these end style

Solution 2

begin mathsize 12px style We space know space cos space left parenthesis 90 minus straight theta right parenthesis space equals space sinθ
space space space space space space space space space space space space space space For space straight theta space equals space 25 to the power of straight o space rightwards double arrow cos thin space left parenthesis 65 to the power of straight o right parenthesis space equals space sin space 25 to the power of straight o space minus space circle enclose 1
space space space space space space space space space space space space space and space sin space squared straight theta space plus space cos squared straight theta space space equals space 1 space minus space circle enclose 2
distance space betweem space given space points space equals space square root of left parenthesis acos 25 minus 0 right parenthesis squared plus left parenthesis 0 minus acos 65 to the power of straight o right parenthesis squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space square root of straight a squared left parenthesis cos squared 25 to the power of straight o plus cos squared 65 to the power of straight o right parenthesis end root
from space circle enclose 1 space and space circle enclose 2
distance space equals space square root of straight a squared end root
space space space space space space space space space space space space space space space space space equals space straight a
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. space space space space end style

Question 3

Error converting from MathML to accessible text.

Solution 3

begin mathsize 12px style distance space equals space square root of left parenthesis straight x minus 3 right parenthesis squared plus left parenthesis 2 plus 6 right parenthesis squared end root
space space space space space space space space space space space rightwards double arrow 10 space equals space square root of left parenthesis straight x minus 3 right parenthesis squared plus 8 squared end root
space space space space space space space space space space space space rightwards double arrow 10 squared space equals space left parenthesis straight x minus 3 right parenthesis squared space plus space 8 squared
space space space space space space space space space space space space rightwards double arrow 100 space equals space 64 space equals space left parenthesis straight x minus 3 right parenthesis squared
space space space space space space space space space space space space rightwards double arrow space left parenthesis straight x minus 3 right parenthesis squared equals space 36
space space space space space space space space space space space space rightwards double arrow straight x minus 3 equals space 6
space space space space space space space space space space space space space rightwards double arrow space box enclose straight x equals 9 end enclose space space or space box enclose straight x equals negative 3 end enclose
But space straight x space is space positive
space space space space space space space space space space space space space space space rightwards double arrow box enclose straight x equals 9 end enclose
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 4

begin mathsize 12px style The space distance space between space the space points space left parenthesis straight a space cos space straight theta space plus space straight b space sin space straight theta comma 0 right parenthesis space and space left parenthesis 0 comma straight a space sin space straight theta space minus space straight b space cos space straight theta right parenthesis space is
left parenthesis straight a right parenthesis space straight a squared space plus space straight b squared
left parenthesis straight b right parenthesis space straight a space plus space straight b
left parenthesis straight c right parenthesis space straight a squared space minus space straight b squared
left parenthesis straight d right parenthesis space square root of straight a squared space plus space straight b squared end root end style

Solution 4

begin mathsize 12px style distance space equals space square root of left parenthesis acosθ space plus space straight b space sinθ space minus 0 right parenthesis squared plus left parenthesis 0 minus asinθ plus bcosθ right parenthesis squared end root
space space space space space space space space space space space space space space space space space space equals square root of left parenthesis straight a squared cos squared straight theta plus straight b squared sin squared straight theta plus 2 absinθcosθ plus straight a squared sin squared straight theta plus straight b squared cos squared straight theta minus space 2 ab space sinθcosθ end root space
space space space space space space space space space space space space space space space space space space equals square root of straight a squared left parenthesis sin squared straight theta plus cos squared straight theta right parenthesis space plus straight b squared space left parenthesis sin squared straight theta plus cos squared straight theta right parenthesis end root
space space space space space space space space space space space space space space space space space space equals square root of straight a squared plus straight b squared end root
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 5

begin mathsize 12px style If space the space distance space between space the space points space left parenthesis 4 comma straight p right parenthesis space and space left parenthesis 1 comma 0 right parenthesis space is space 5 comma space then space straight p equals
left parenthesis straight a right parenthesis thin space space plus-or-minus 4
left parenthesis straight b right parenthesis space 4
left parenthesis straight c right parenthesis space minus 4
left parenthesis straight d right parenthesis space 0 end style

Solution 5

begin mathsize 12px style distance space equals space square root of left parenthesis 4 minus 1 right parenthesis squared plus left parenthesis straight p minus 0 right parenthesis squared end root
space space space space space space space space rightwards double arrow space 5 space equals space square root of 3 squared plus straight p squared end root
space space space space space space space rightwards double arrow 25 space equals space straight q space plus space straight p squared
space space space space space space space rightwards double arrow straight p squared space equals space 16
space space space space space space space rightwards double arrow straight p space equals space plus-or-minus 4
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 6

A line Segement is of length 10 units. If the coordinates of its one end are (2,3) and the abscissa of the other end is 10, then its ordinate is

(a) 9,6

(b) 3,-9

(c) -3,9

(d) 9,-6

Solution 6

begin mathsize 12px style Let space the space abscissa space be space straight x.
Then space distance space between space points space left parenthesis 2 comma negative 3 right parenthesis space and space left parenthesis 10 comma straight x right parenthesis space is
equals space square root of left parenthesis 2 minus 10 right parenthesis squared plus left parenthesis negative 3 minus straight x right parenthesis squared end root
equals space square root of 8 squared plus left parenthesis straight x plus 3 right parenthesis squared end root
space 10 equals square root of 64 space plus space left parenthesis straight x plus 3 right parenthesis squared end root
rightwards double arrow space 100 space equals space 64 space plus space open parentheses straight x plus 3 close parentheses squared
rightwards double arrow space left parenthesis straight x plus 3 right parenthesis squared space equals space 36
rightwards double arrow straight x space plus space 3 space equals space 6 space or space straight x space plus space 3 space equals space minus 6
rightwards double arrow straight x space equals space 3 space or space straight x equals space minus 9
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 7

begin mathsize 12px style The space perimeter space of space the space triangle space formed space by space the space points space left parenthesis 0 comma 0 right parenthesis comma left parenthesis 1 comma 0 right parenthesis space and space left parenthesis 0 comma 1 right parenthesis space is
left parenthesis straight a right parenthesis space 1 plus-or-minus space square root of 2
left parenthesis straight b right parenthesis space square root of 2 space plus space 1 end root
left parenthesis straight c right parenthesis space 3
left parenthesis straight d right parenthesis space 2 space plus space square root of 2

end style

Solution 7

begin mathsize 12px style AB equals space square root of left parenthesis 0 minus 1 right parenthesis squared plus left parenthesis 0 minus 0 right parenthesis squared end root
space space space space space space space space space space space space space space space space space space space space space equals space 1
BC space equals space square root of left parenthesis 1 minus 0 right parenthesis squared plus left parenthesis 0 minus 1 right parenthesis squared end root
space space space space space space space space space space space space space space space space space space space space equals square root of 1 plus 1 end root
space space space space space space space space space space space space space space space space space space space space space equals square root of 2
CA space equals space square root of left parenthesis 0 minus 0 right parenthesis 2 space plus space left parenthesis 1 minus 0 right parenthesis squared end root
space space space space space space space space space space space space space space space space space space space space space space equals space 1
Perimeter equals AB plus BC plus AC equals 2 plus square root of 2 end style

So, the correct option is (d).

Question 8

begin mathsize 12px style If space straight A space left parenthesis 2 comma 2 right parenthesis comma thin space straight B left parenthesis negative 4 comma negative 4 right parenthesis space and space straight C left parenthesis 5 comma negative 8 right parenthesis space are space the space vertices space of space straight a space triangle comma space then space the space length
of space the space median space through space verte space straight C space is
left parenthesis straight a right parenthesis space square root of 65
left parenthesis straight b right parenthesis space square root of 117
left parenthesis straight c right parenthesis square root of 85
left parenthesis straight d right parenthesis square root of 113 end style

Solution 8

begin mathsize 12px style straight D space is space mid space point space of space straight A space and space straight B
Hence space Coordinates space of space straight D space are
open parentheses fraction numerator 2 plus left parenthesis negative 4 right parenthesis over denominator 2 end fraction comma space fraction numerator 2 plus open parentheses negative 4 close parentheses over denominator 2 end fraction close parentheses
open parentheses fraction numerator negative 2 over denominator 2 end fraction comma fraction numerator negative 2 over denominator 2 end fraction close parentheses
left parenthesis negative 1 comma negative 1 right parenthesis
Length space of space median space CD space equals space square root of left parenthesis 5 plus 1 right parenthesis squared plus left parenthesis negative 8 plus 1 right parenthesis squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of 6 squared plus 7 squared end root space equals space square root of 36 plus 49 end root space equals space square root of 85 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
space So comma space the space correct space option space is space left parenthesis straight c right parenthesis. space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end style

Question 9

begin mathsize 12px style If space three space points space left parenthesis 0 comma 0 right parenthesis comma space left parenthesis 3 comma square root of 3 right parenthesis end root space and space left parenthesis 3 comma straight lambda right parenthesis space form space and space equilateral space triangle comma space then space straight lambda space equals
left parenthesis straight a right parenthesis space 2
left parenthesis straight b right parenthesis space minus 3
left parenthesis straight c right parenthesis space minus space 4
left parenthesis straight d right parenthesis space None space of space these space end style

Solution 9

 begin mathsize 12px style All space sides space of space straight a space equilateral space traingle space are space equal
space space space space space space space space space space space space space space AC space equals space square root of left parenthesis 3 minus 0 right parenthesis squared plus left parenthesis square root of 3 minus 0 right parenthesis squared end root
space space space space space space space space space space space space space space space space space space space space space equals space square root of 9 plus 3 end root
space space space space space space space space space space space space space space space space space space space space space space equals space square root of 12
space space space space space space space space space space space space space space space AC space equals space 2 square root of 3
AB space equals space square root of left parenthesis 3 minus 0 right parenthesis squared left parenthesis straight lambda minus 0 right parenthesis squared end root
space space space space space space space equals space square root of straight lambda squared plus 9 end root
But space AB space equals space AC
space space rightwards double arrow square root of straight lambda squared plus 9 end root equals square root of 12
space space space rightwards double arrow straight lambda squared plus 9 space equals space 12
space space space rightwards double arrow straight lambda squared space equals space 3
space space space space rightwards double arrow straight lambda space equals space plus-or-minus square root of 3
But space if space straight lambda space equals space square root of 3 space then space straight B space and space straight C space are space same
Hence space box enclose straight lambda space equals space minus square root of 3 end enclose
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 10

begin mathsize 12px style If space the space points space left parenthesis straight k comma 2 straight k right parenthesis comma space left parenthesis 3 straight k comma 3 straight k right parenthesis space and space left parenthesis 3 comma 1 right parenthesis space are space collinear comma space then space straight k
left parenthesis straight a right parenthesis space 1 third
left parenthesis straight b right parenthesis space minus 1 third
left parenthesis straight c right parenthesis space 2 over 3
left parenthesis straight d right parenthesis space minus space 2 over 3 end style

Solution 10

begin mathsize 12px style If space there space points space are space collinear space then space co minus ordinate space if space third space point space must
satisfy space the space straight space line space passing space through space the space first space two space points
rightwards double arrow Straight space line space passing space through space left parenthesis straight k comma 2 straight k right parenthesis space and space left parenthesis 3 straight k comma 3 straight k right parenthesis space is space fraction numerator straight y minus 2 straight k over denominator straight x minus straight k end fraction equals fraction numerator 3 straight k minus 2 straight k over denominator 3 straight k minus straight k end fraction minus space circle enclose 1
The space point space left parenthesis 3 comma 1 right parenthesis space satisfies space eq space circle enclose 1
rightwards double arrow fraction numerator 1 minus 2 straight k over denominator 3 minus straight k end fraction equals fraction numerator straight k over denominator 2 straight k end fraction
rightwards double arrow fraction numerator 1 minus 2 straight k over denominator 3 minus straight k end fraction equals 1 half
rightwards double arrow 2 minus 4 space straight k equals 3 minus straight k
rightwards double arrow negative 1 space equals space 3 straight k
rightwards double arrow straight k space equals space minus 1 third
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Chapter 6 - Co-ordinate Geometry Exercise 6.64

Question 1

begin mathsize 12px style The space coordinates space of space the space point space on space straight x minus axis space which space ar space equidistant space from space the space points space
left parenthesis negative 3 comma 4 right parenthesis space and space left parenthesis 2 comma 5 right parenthesis space are
left parenthesis straight a right parenthesis space left parenthesis 20 comma 0 right parenthesis
left parenthesis straight b right parenthesis space left parenthesis negative 23 comma 0 right parenthesis
left parenthesis straight c right parenthesis space open parentheses 4 over 5 comma 0 close parentheses
left parenthesis straight d right parenthesis space None space of space these end style

Solution 1

begin mathsize 12px style Given space that space point space is space on space straight x minus axis
Hence space let space point space is space left parenthesis straight x comma 0 right parenthesis
space distance space from space space left parenthesis negative 3 comma 4 right parenthesis space equals space square root of left parenthesis straight x plus 3 right parenthesis squared plus left parenthesis 0 minus 4 right parenthesis squared end root space space space space space space space space space space space space space space minus space circle enclose 1
distance space from space left parenthesis 2 comma 5 right parenthesis space equals space square root of left parenthesis straight x minus 2 right parenthesis squared plus left parenthesis 0 minus 5 right parenthesis squared end root space space space space space space space space space space space space space space space space space space space space minus circle enclose 2
But space circle enclose 1 space and space circle enclose 2 space end enclose space are space equal
rightwards double arrow square root of left parenthesis straight x plus 3 right parenthesis squared plus 16 space end root space equals space square root of left parenthesis straight x minus 2 right parenthesis squared plus 25 end root
rightwards double arrow space left parenthesis straight x plus 3 right parenthesis squared space plus space 16 space equals space left parenthesis straight x minus 2 right parenthesis squared space plus space 25
rightwards double arrow straight x squared space plus space 9 space plus space 6 straight x space plus space 16 space equals space straight x squared space minus space 4 straight x space plus space 4 space plus space 25
rightwards double arrow 10 straight x space equals space 4
rightwards double arrow straight x space equals 2 over 5
rightwards double arrow space Coordinates space open parentheses 2 over 5 comma 0 close parentheses
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 2

If (-1,2),(2,-1) and (3,1) are any three vertices of a parallelogram, then

(a) a = 2, b = 0

(b) a = -2, b = 0

(c) a = -2, b = 6

(d) a = 6, b = 2

Solution 2

begin mathsize 12px style If space ABCD space is space straight a space paralelogram space then space AB space II space CD space and space AD space II space BC
If space AB space II space CD
Slope space of space AB equals space Slope space of space CD
rightwards double arrow fraction numerator 2 minus left parenthesis negative 1 right parenthesis over denominator negative 1 minus left parenthesis 2 right parenthesis end fraction equals fraction numerator straight b minus 1 over denominator straight a minus 3 end fraction
rightwards double arrow fraction numerator 3 over denominator negative 3 end fraction space equals space fraction numerator straight b minus 1 over denominator straight a minus 3 end fraction
rightwards double arrow space minus 1 space equals space fraction numerator straight b minus 1 over denominator straight a minus 3 end fraction
rightwards double arrow negative straight a plus 3 space equals space straight b minus 1
rightwards double arrow box enclose straight a plus straight b space equals space 4 end enclose space minus space circle enclose 1
And space AD space II space BC
slope space of space AD space equals space slope space of space BC
rightwards double arrow fraction numerator 2 minus straight b over denominator negative 1 minus a end fraction space equals space fraction numerator negative 1 minus 1 over denominator 2 minus 3 end fraction
rightwards double arrow fraction numerator straight b minus 2 over denominator straight a plus 1 end fraction space equals space 2 over 1
rightwards double arrow straight b space minus 2 space equals space 2 straight a plus 2
rightwards double arrow box enclose 2 straight a minus straight b space equals space minus 4 end enclose space minus space circle enclose 2
circle enclose 1 space plus space circle enclose 2
rightwards double arrow space 3 straight a space equals space 0
box enclose straight a space equals space 0 end enclose space and space box enclose straight b space equals space 4 end enclose space space space space space space space space space space space space space end style

 

Note: The answer does not match the options.

Question 3

If A (5,3), B(11,-5) and P(12,y) are the vertices of a right triangle right angled at P, then y = 

(a) - 2,4

(b) -2 ,4

(c) 2, -4

(d) 2,4

Solution 3

begin mathsize 12px style AB squared space equals space AP squared space plus space left parenthesis PB right parenthesis to the power of 2 space end exponent minus space circle enclose 1
AB space equals space square root of left parenthesis 11 minus 5 right parenthesis squared space plus space left parenthesis negative 5 minus 3 right parenthesis squared end root
space space space space space space space equals space square root of 6 squared space plus space 8 squared end root
AB space equals space 10 space space space space space space space space minus space circle enclose 2
AP space equals space square root of left parenthesis 12 minus 5 right parenthesis squared left parenthesis straight y minus 3 right parenthesis squared end root
AP space equals space square root of 7 squared plus left parenthesis straight y minus 3 right parenthesis squared end root space minus space circle enclose 3
BP space equals space square root of left parenthesis 12 minus 11 right parenthesis squared plus left parenthesis straight y plus 5 right parenthesis squared end root
space space space space space space space equals space square root of 1 plus open parentheses straight y plus 5 close parentheses squared end root space space minus space circle enclose 4
From space circle enclose 1 comma circle enclose 2 comma circle enclose 3 comma circle enclose 4
rightwards double arrow space 10 squared space equals space open parentheses square root of 7 squared plus open parentheses straight y minus 3 close parentheses squared end root close parentheses space plus space open parentheses square root of 1 plus open parentheses straight y plus 5 close parentheses squared end root close parentheses squared
rightwards double arrow space 100 space equals space 49 plus space left parenthesis straight y minus 3 right parenthesis squared space plus space 1 space plus space open parentheses straight y plus 5 close parentheses squared
rightwards double arrow space 100 space equals space 50 space plus space straight y squared space minus space 6 straight y space plus space 9 space plus space straight y squared space plus space 10 straight y space plus space 25
rightwards double arrow 2 straight y squared space minus space 4 straight y space minus space 16 space equals space 0
rightwards double arrow straight y squared space plus space 2 straight y space minus 8 space equals space 0
rightwards double arrow straight y squared space plus space 4 straight y space minus space 2 straight y space minus space 8 space equals space 0
rightwards double arrow straight y space left parenthesis straight y plus 4 right parenthesis space minus space 2 space left parenthesis straight y plus 4 right parenthesis space equals space 0
rightwards double arrow straight y space equals space 2 comma space minus 4
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 4

The area of the triangle formed by (a,b+c), (b,c+a) and (c,a+b) is 

(a)  a+b+c

(b) abc

(c) (a+b+c)2

(d) 0

Solution 4

begin mathsize 12px style We space know space know space if space left parenthesis straight x comma straight y right parenthesis comma space left parenthesis straight x subscript 2 comma straight y subscript 2 right parenthesis space and space left parenthesis straight x subscript 3 comma straight y subscript 3 right parenthesis space are space co minus ordinate space of space any space triangle space then
area space of space triangle space equals space open vertical bar fraction numerator straight x subscript 1 comma left parenthesis straight y subscript 2 minus straight y subscript 3 right parenthesis space plus space straight x subscript 2 left parenthesis straight y subscript 3 minus straight y subscript 1 right parenthesis plus straight x subscript 3 left parenthesis straight y subscript 1 minus straight y subscript 2 right parenthesis over denominator 2 end fraction close vertical bar
rightwards double arrow space Coordinates space are space left parenthesis straight a comma straight b plus straight c right parenthesis comma space left parenthesis straight b comma straight c plus straight a right parenthesis comma space left parenthesis straight c comma straight a plus straight b right parenthesis
rightwards double arrow area space of space triangle space equals space open vertical bar fraction numerator straight a left parenthesis straight c plus straight a right parenthesis minus left parenthesis straight a plus straight b right parenthesis plus straight b left parenthesis straight a plus straight b right parenthesis minus left parenthesis straight b plus straight c right parenthesis plus straight c left parenthesis left parenthesis straight b plus straight c right parenthesis minus left parenthesis straight c plus straight a right parenthesis right parenthesis over denominator 2 end fraction close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open vertical bar fraction numerator straight a left parenthesis straight c minus straight b right parenthesis plus straight b left parenthesis straight a minus straight c right parenthesis plus straight c left parenthesis straight b minus straight a right parenthesis over denominator 2 end fraction close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open vertical bar fraction numerator ac minus ab space plus space ab space minus space bc space plus space bc space minus space ac over denominator 2 end fraction close vertical bar
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 5

If (x,2),(-3,-4) and (7,-5) are collinear, then x = 

(a) 60

(b) 63

(c) -63

(d) -60

Solution 5

begin mathsize 12px style If space 3 space points space are space collinear space comma space then space area space of space triangle space formed space by space theses space 3 space ponts
must space be space 0.
rightwards double arrow space area space equals space open vertical bar fraction numerator straight x left parenthesis negative 4 space plus space 5 right parenthesis space minus space 3 space left parenthesis negative 5 space minus 2 right parenthesis space plus space 7 space left parenthesis 2 plus 4 right parenthesis over denominator 2 end fraction close vertical bar equals 0
rightwards double arrow straight x space plus space 21 space plus space 42 space equals space 0
rightwards double arrow space straight x space equals space minus space 63
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 6

begin mathsize 12px style If space points space left parenthesis straight t comma 2 straight t right parenthesis comma space left parenthesis negative 2 comma 6 right parenthesis space and space left parenthesis 3 comma 1 right parenthesis space are space collinear space then space straight t space equals space
left parenthesis straight a right parenthesis space 3 over 4
left parenthesis straight b right parenthesis space 4 over 3
left parenthesis straight c right parenthesis space 5 over 3
left parenthesis straight d right parenthesis space 3 over 5 end style

Solution 6

begin mathsize 12px style area space equals space open vertical bar fraction numerator straight t left parenthesis 6 minus 1 right parenthesis minus 2 left parenthesis 1 minus 2 straight t right parenthesis plus 3 left parenthesis 2 straight t minus 6 right parenthesis over denominator 2 end fraction close vertical bar equals 0
rightwards double arrow 5 straight t space plus space 4 straight t space minus space 2 space plus space 6 straight t space minus space 18 space equals space 0
rightwards double arrow space 15 straight t space equals space 20
rightwards double arrow straight t space equals space 4 over 3
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 7

begin mathsize 12px style If space the space area space of space the space triangle space formed space by space the space points space left parenthesis straight x comma 2 straight x right parenthesis comma space left parenthesis negative 2 comma 6 right parenthesis space and space left parenthesis 3 comma 1 right parenthesis space is space 5 space square space units comma space then space straight x space equals space
left parenthesis straight a right parenthesis space 2 over 3
left parenthesis straight b right parenthesis space 3 over 5
left parenthesis straight c right parenthesis space 3
left parenthesis straight d right parenthesis space 5 end style

Solution 7

begin mathsize 12px style area space equals space open vertical bar fraction numerator straight x left parenthesis 6 minus 1 right parenthesis minus 2 left parenthesis 1 minus 2 straight x right parenthesis plus 3 left parenthesis 2 straight x minus 6 right parenthesis over denominator 2 end fraction close vertical bar
rightwards double arrow space 5 space equals space open vertical bar fraction numerator 5 straight x space plus space 4 straight x minus 2 space plus space 6 straight x space minus space 18 over denominator 2 end fraction close vertical bar
rightwards double arrow 15 straight x space minus space 20 space equals space 10 space or space 15 straight x space minus space 20 space equals space minus 10
rightwards double arrow 15 straight x equals space 30 space space or space 15 straight x space equals space 10
rightwards double arrow straight x space equals space 2 space or space straight x space equals space 2 over 3
So space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 8

begin mathsize 12px style If space points space left parenthesis straight a comma 0 right parenthesis left parenthesis 0 comma straight b right parenthesis space and space left parenthesis 1 comma 1 right parenthesis space are space collinear comma space then space 1 over straight a plus 1 over straight b equals
left parenthesis straight a right parenthesis space 1
left parenthesis straight b right parenthesis space 2
left parenthesis straight c right parenthesis space 0
left parenthesis straight d right parenthesis space minus 1 end style

Solution 8

begin mathsize 12px style area space equals space open vertical bar fraction numerator straight a left parenthesis straight b minus 1 right parenthesis plus 0 left parenthesis 1 minus 0 right parenthesis plus 1 left parenthesis 0 minus straight b right parenthesis over denominator 2 end fraction close vertical bar
rightwards double arrow 0 space equals space open vertical bar fraction numerator ab space minus space straight a space minus space straight b over denominator 2 end fraction close vertical bar
rightwards double arrow ab space equals space straight a space plus space straight b
rightwards double arrow 1 space equals space fraction numerator straight a plus straight b over denominator ab end fraction
rightwards double arrow 1 over straight a plus 1 over straight b equals 1
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 9

begin mathsize 12px style If space the space centriod space of space straight a space triangle space is space space left parenthesis 1 comma 4 right parenthesis space and space two space if space its space vertices space are space left parenthesis 4 comma negative 3 right parenthesis space and space left parenthesis negative 9 comma 7 right parenthesis comma space then
the space area space of space the space triangle space is
left parenthesis straight a right parenthesis space 183 space sq. units
left parenthesis straight b right parenthesis space 183 over 2 sq. units
left parenthesis straight c right parenthesis space 366 space sq. units
left parenthesis straight d right parenthesis space 183 over 4 sq. units end style

Solution 9

begin mathsize 12px style We space know space if space left parenthesis straight x comma straight y right parenthesis comma space left parenthesis left parenthesis straight x subscript 2 comma straight y subscript 2 right parenthesis comma space straight x subscript 3 comma straight y subscript 3 right parenthesis space are space co minus ordinates space of space triangle space then space coordinate space of space centriod
is space open parentheses fraction numerator straight x subscript 1 plus straight x subscript 2 plus straight x subscript 3 over denominator 3 end fraction comma fraction numerator straight y subscript 1 plus straight y subscript 2 plus straight y subscript 3 over denominator 3 end fraction close parentheses space minus space circle enclose 1
rightwards double arrow space Two space vertices space of space of space triangle space are space left parenthesis 4 comma negative 3 right parenthesis space and space left parenthesis negative 9 comma 7 right parenthesis
Let space third space coordinate space be space left parenthesis straight a comma straight b right parenthesis
Given space centroid space is space left parenthesis 1 comma 4 right parenthesis
Now space form space circle enclose 1
fraction numerator straight a plus 4 minus 9 over denominator 3 end fraction equals 1 space and space fraction numerator straight b minus 3 plus 7 over denominator 3 end fraction equals 4
rightwards double arrow straight a space minus space 5 space equals space 3 space space space space and space space space straight b plus 4 equals 12
box enclose straight a space equals space 8 end enclose space space space space and space space space box enclose straight b equals 8 end enclose
All space co minus ordinates space are space left parenthesis 8 comma 8 right parenthesis comma space left parenthesis 4 comma negative 3 right parenthesis space and space left parenthesis straight a minus 9 comma 7 right parenthesis
area space equals space open double vertical bar fraction numerator 8 left parenthesis negative 3 minus 7 right parenthesis space plus space 4 space left parenthesis 7 space minus space 8 right parenthesis space minus 9 left parenthesis 8 plus 3 over denominator 2 end fraction close double vertical bar
space space space space space space space space space equals open vertical bar fraction numerator negative 80 space minus space 4 space minus space 99 over denominator 2 end fraction close vertical bar
space space space space space space space space equals open vertical bar fraction numerator negative 183 over denominator 2 end fraction close vertical bar
space space space space space space equals space 183 over 2 space square space units
So space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 10

The line segment joining points (-3,4), and (1,-2) is divided by y - axis in the ratio

(a) 1 : 3

(b) 2 : 3

(c) 3 : 1

(d) 2: 3

Solution 10

begin mathsize 12px style Let space straight y minus axis space divde space the space space segment space in space straight m colon straight n
straight x space co minus ordinate space of space straight P
rightwards double arrow 0 equals space fraction numerator straight m cross times 1 plus straight n cross times left parenthesis negative 3 right parenthesis over denominator straight m plus straight n end fraction
rightwards double arrow 0 space equals space straight m space minus space 3 straight n
rightwards double arrow straight m over straight n space space equals space 3 colon 1
So comma space the space correct space option space left parenthesis straight c right parenthesis. end style

Question 11

The ratio in which (4,5) divides the join of (2,3) and (7,8) is 

(a) -2 : 3

(b) -3 : 2 

(c) 3 : 2

(d) 2 : 3

Solution 11

begin mathsize 12px style We space know space if space any space line space segment space AB space is space decide space by space straight C space in space straight m colon straight n space rather space then
straight x space equals space fraction numerator mx subscript 2 space plus space nx subscript 1 over denominator straight m plus straight n end fraction
straight y space equals space fraction numerator my subscript 2 space plus space ny subscript 1 over denominator straight m plus straight n end fraction
Let space left parenthesis 4 comma 5 right parenthesis decides space the space join space left parenthesis 2 comma 3 right parenthesis space and space left parenthesis 7 comma 8 right parenthesis space in space straight m colon straight n space then space 4 space equals space fraction numerator straight m cross times 7 plus 2 cross times straight n over denominator straight m plus straight n end fraction
rightwards double arrow 4 straight m space plus space 4 straight n space equals space 7 straight m space plus space 2 straight n
rightwards double arrow space 2 straight n space equals space 3 straight m
rightwards double arrow straight m over straight n space equals space 2 over 3
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 12

The ratio in which the x-axis divides the segment joining (3,6) and (12,-3) is

(a) 2:1

(b) 1 :2

(c) -2 : 1

(d) 1 : -2 

Solution 12

begin mathsize 12px style 0 space equals space fraction numerator straight m cross times left parenthesis negative 3 right parenthesis space plus space straight n cross times 6 over denominator straight m plus straight n end fraction
rightwards double arrow negative 3 straight m space plus space 6 straight n space equals space 0
rightwards double arrow 3 straight m space equals space 6 straight n
rightwards double arrow straight m over straight n space equals space 2 space colon space 1
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 13

If the centrroid  of the triangle formed by the points (a,b),(b,c) and (c,a) is at the origin, then a3 + b3 + c3

(a) abc

(b) 0

(c) a+b+c

(d) 3abc

Solution 13

begin mathsize 12px style Co minus ordinate space of space centriod space equals space open parentheses fraction numerator straight a plus straight b plus straight c over denominator 3 end fraction comma fraction numerator straight a plus straight b plus straight c over denominator 3 end fraction close parentheses
but space given space that space centroid space at space origin space Hence space box enclose straight a plus straight b plus straight c space equals space 0 end enclose space minus space circle enclose 1
We space know space straight a cubed space plus space straight b cubed space plus space straight c cubed space minus space 3 abc space equals space left parenthesis straight a plus straight b plus straight c right parenthesis left parenthesis straight a squared plus straight b squared plus straight c squared space minus space ab space minus space bc space minus space ca right parenthesis space minus space circle enclose 2
from space circle enclose 1 space and space circle enclose 2
straight a cubed space plus space straight b cubed space plus space straight c cubed space equals space 3 abc
Hence comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 14

If points (1, 2), (-5, 6) and (a, -2) are collinear, then a =

(a) -3

(b) 7

(c) 2

(d) -2

Solution 14

begin mathsize 12px style If space points space are space collinear space then space area space of space triangle space is space zero
area space equals space open vertical bar fraction numerator 1 open parentheses 6 plus 2 close parentheses minus 5 open parentheses negative 2 minus 2 close parentheses plus straight a open parentheses 2 minus 6 close parentheses over denominator 2 end fraction close vertical bar
0 space equals open vertical bar fraction numerator 8 plus 20 minus 4 straight a over denominator 2 end fraction close vertical bar
rightwards double arrow space 28 space minus space 4 straight a space equals space 0
rightwards double arrow space straight a space equals space 7
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 15

If the centroid of the triangle formed by (7, x), (y, -6) and (9, 10) is at (6, 3), then (x, y) =

(a) (4, 5)

(b) (5, 4)

(c) (-5, -2)

(d) (5, 2)

Solution 15

begin mathsize 12px style Coordinate space of space centroid space equals space open parentheses fraction numerator 7 plus straight y plus 9 over denominator 3 end fraction comma space fraction numerator straight x minus 6 plus 10 over denominator 3 end fraction close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses fraction numerator straight y plus 16 over denominator 3 end fraction comma space fraction numerator straight x plus 4 over denominator 3 end fraction close parentheses
Given space left parenthesis 6 comma space 3 right parenthesis
Hence space space space space 6 space equals space fraction numerator straight y plus 16 over denominator 3 end fraction space space space space space space space space space space and space space space 3 space equals space fraction numerator straight x plus 4 over denominator 3 end fraction
space space space space space space space space space space space space rightwards double arrow space 18 space equals space straight y plus 16 space space space space space space space space space space space space space space space space space space space 9 space equals space straight x plus 4
space space space space space space space space space space space space rightwards double arrow space straight y space equals space 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight x space equals space 5
left parenthesis straight x comma space straight y right parenthesis space equals space left parenthesis 5 comma space 2 right parenthesis
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Chapter 6 - Co-ordinate Geometry Exercise 6.65

Question 1

The distance of the point (4, 7) from the x-axis is

(a) 4

(b) 7

(c) 11

(d) begin mathsize 12px style square root of 65 end style

Solution 1

We know, distance of point (x, y) from x-axis is y.

Hence, distance of point (4, 7) from x-axis is 7.

Hence, correct option is (b).

Question 2

The distance of the point (4, 7) from the y-axis is

(a) 4

(b) 7

(c) 11

(d) begin mathsize 12px style square root of 65 end style

Solution 2

We know distance of point (x, y) from y-axis is x.

Hence distance of point (4, 7) from y-axis is 4.

Hence, correct option is (a).

Question 3

If P is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point Q on OY such that OP = OQ, are

(a) (0, 3)

(b) (3, 0)

(c) (0, 0)

(d) (0, -3)

Solution 3

Given P is a point on x-axis

Hence P = (x, 0)

Distance from the origin is 3

Hence P = (3, 0)

Given Q is a point on y-axis

So Q is (0, y)

Given that OP = OQ

implies OQ = 3

Distance of Q from the origin is 3

Hence y = 3

implies Q = (0, 3)

Hence, correct option is (a).

Question 4

If the point (x, 4) lies on a circle whose centre is at the origin and radius is 5, then x =

(a) ±5

(b) ±3

(c) 0

(d) ±4

Solution 4

begin mathsize 12px style Centre space equals space left parenthesis 0 comma space 0 right parenthesis
point space equals space left parenthesis straight x comma space 4 right parenthesis
radius space equals space 5
Distance space between space centre space and space the space point space is space equal space to space the space radius
rightwards double arrow square root of left parenthesis straight x minus 0 right parenthesis squared plus open parentheses 4 minus 0 close parentheses squared end root space equals space 5
rightwards double arrow space straight x squared space plus space 4 squared space equals space 25
rightwards double arrow space straight x squared space equals space 9
rightwards double arrow straight x equals plus-or-minus 3
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 5

If the point P (x, y) is equidistant from A(5, 1) and B (-1, 5), then

(a) 5x = y

(b) x = 5y

(c) 3x = 2y

(d) 2x = 3y

Solution 5

begin mathsize 12px style AP space equals space square root of open parentheses straight x minus 5 close parentheses squared plus open parentheses straight y minus 1 close parentheses squared end root
BP space equals space square root of open parentheses straight x plus 1 close parentheses squared plus open parentheses straight y minus 5 close parentheses squared end root
Given space AP space equals space BP
rightwards double arrow square root of open parentheses straight x minus 5 close parentheses squared plus open parentheses straight y minus 1 close parentheses squared end root space equals space square root of open parentheses straight x plus 1 close parentheses squared plus open parentheses straight y minus 5 close parentheses squared end root
rightwards double arrow open parentheses straight x minus 5 close parentheses squared plus open parentheses straight y minus 1 close parentheses squared space equals space open parentheses straight x plus 1 close parentheses squared plus open parentheses straight y minus 5 close parentheses squared
rightwards double arrow space straight x squared space minus space 10 straight x space plus space 25 space plus space straight y squared space minus space 2 straight y space plus space 1 space equals space straight x squared space plus space 2 straight x space plus space 1 space plus space straight y squared space minus space 10 straight y space plus space 25
rightwards double arrow negative 10 straight x space minus space 2 straight y space equals space 2 straight x space minus space 10 straight y
rightwards double arrow space 8 straight y space equals space 12 straight x
rightwards double arrow space 3 straight x space equals space 2 straight y
Hence comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 6

If points A (5, p), B (1, 5), C (2, 1) and D (6, 2) form a square ABCD, then p =

(a) 7

(b) 3

(c) 6

(d) 8

Solution 6

begin mathsize 12px style All space sides space of space square space are space equal
AB space equals BC space equals space CD space equals space AD
rightwards double arrow space AB space equals space BC
rightwards double arrow space square root of open parentheses 5 minus 1 close parentheses squared plus open parentheses straight P minus 5 close parentheses squared end root space equals space square root of open parentheses 1 minus 2 close parentheses squared plus open parentheses 5 minus 1 close parentheses squared end root
rightwards double arrow space 4 squared plus space open parentheses straight P minus 5 close parentheses squared space space space space space space space space space space space space space space equals space 1 squared space plus space 4 squared
rightwards double arrow space open parentheses straight p minus 5 close parentheses squared space equals space 1 squared
rightwards double arrow space straight p minus 5 space equals space 1 space space space space space space space space space space space space space space space space space space space space space space space space space or space straight p minus 5 equals negative 1
rightwards double arrow space straight p equals 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight p space equals space 4
If space we space take space straight p equals 4 comma
Now comma space BC space equals space space space space square root of open parentheses 1 minus 2 close parentheses squared plus open parentheses 5 minus 1 close parentheses squared end root equals square root of 1 squared space plus space 4 squared end root equals square root of 17
space AD equals square root of open parentheses 5 minus 6 close parentheses squared plus open parentheses 4 minus 2 close parentheses squared end root equals square root of 1 space plus space 4 end root equals square root of 5
So comma space the space sides space of space the space square space are space not space equal space if space straight p equals 4
Hence comma space straight p space equals space 6
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 7

begin mathsize 12px style The space coordinates space of space the space circumcentre space of space the space triangle space formed space by space the space points space straight O space left parenthesis 0 comma 0 right parenthesis comma space straight A space left parenthesis straight a comma space 0 right parenthesis space and space straight B space left parenthesis 0 comma space straight b right parenthesis space are
left parenthesis straight a right parenthesis space left parenthesis straight a comma space straight b right parenthesis
left parenthesis straight b right parenthesis space open parentheses straight a over 2 comma space straight b over 2 close parentheses
left parenthesis straight c right parenthesis space open parentheses straight b over 2 comma space straight a over 2 close parentheses
left parenthesis straight d right parenthesis space left parenthesis straight b comma space straight a right parenthesis end style

Solution 7

begin mathsize 12px style Coordinates space of space the space circumcentre space of space the space triangle space formed space by space left parenthesis straight x subscript 1 comma space straight y subscript 1 right parenthesis space left parenthesis straight x subscript 2 comma space straight y subscript 2 right parenthesis space left parenthesis straight x subscript 3 comma space straight y subscript 3 right parenthesis space is
the space point space of space intersection space of space the space perpendicular space bisectors space of space the space sides.
straight D space is space the space mid minus point space of space OA space equals space open parentheses fraction numerator 0 plus straight a over denominator 2 end fraction comma 0 close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses straight a over 2 comma 0 close parentheses
straight E space is space the space mid minus point space of space OB space equals open parentheses 0 comma space fraction numerator 0 plus straight b over denominator 2 end fraction close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses 0 comma space straight b over 2 close parentheses space space space
Hence comma space straight F space is space open parentheses straight a over 2 comma space straight b over 2 close parentheses space
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 8

The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (-3, 4) are

(a) (0, 2)

(b) (3, 0)

(c) (0, 3)

(d) (2, 0)

Solution 8

begin mathsize 12px style Coordinates space of space straight a space point space on space the space straight x minus axis space is space left parenthesis straight x comma space 0 right parenthesis
Mid minus point space of space left parenthesis 7 comma space 6 right parenthesis space and space left parenthesis negative 3 comma space 4 right parenthesis space is
open parentheses fraction numerator 7 plus 3 over denominator 2 end fraction comma space fraction numerator 6 minus 4 over denominator 2 end fraction close parentheses equals open parentheses 5 comma space 2 close parentheses
Line space passing space through space left parenthesis 7 comma space 6 right parenthesis space and space left parenthesis negative 3 comma space 4 right parenthesis space is space perpendicular space to space the space line space passing space through space left parenthesis straight x comma space 0 right parenthesis space and space left parenthesis 2 comma space 5 right parenthesis
straight L 1 space rightwards double arrow space Slope space of space the space line space passing space through space left parenthesis 7 comma space 6 right parenthesis space and space left parenthesis negative 3 comma space 4 right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 6 minus 4 over denominator 7 plus 3 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 over 10 equals 1 fifth
straight L 2 space rightwards double arrow Slope space of space the space line space passing space through space left parenthesis straight x comma space 0 right parenthesis space and space left parenthesis 2 comma space 5 right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 5 minus 0 over denominator 2 minus straight x end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 5 over denominator 2 minus straight x end fraction
open parentheses Slope space of space straight L 1 close parentheses space cross times space open parentheses Slope space of space straight L 2 close parentheses space equals space minus 1
rightwards double arrow 1 fifth cross times fraction numerator 5 over denominator 2 minus straight x end fraction equals negative 1
rightwards double arrow space 1 space equals space straight x space minus space 2
rightwards double arrow space straight x space equals space 3
So comma space the space coordinates space are space left parenthesis 3 comma space 0 right parenthesis.
Hence comma space correct space option space are space left parenthesis straight b right parenthesis. end style

Question 9

If the centroid of the triangle formed by the points (3, -5), (-7, 4), (10, -k) is at the point (k, - 1), then k =

(a) 3

(b) 1

(c) 2

(d) 4

Solution 9

begin mathsize 12px style Coordinates space of space the space centroid space equals space open parentheses fraction numerator 3 minus 7 plus 10 over denominator 3 end fraction comma fraction numerator negative 5 plus 4 minus straight k over denominator 3 end fraction close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses 2 comma space fraction numerator negative 1 minus straight k over denominator 3 end fraction close parentheses.... open parentheses 1 close parentheses
Given space centroid space is space at space the space point space open parentheses straight k comma space minus 1 close parentheses space space space.... open parentheses 2 close parentheses
rightwards double arrow space straight k space equals space 2..... space left parenthesis On space compare space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses right parenthesis
Hence comma space correct space option space is space open parentheses straight c close parentheses. end style

Question 10

If (-2, 1) is the Centroid of the triangle having its vertices at (x, 2), (10, -2), (-8, y), then x, y satisfy the relation

a. 3x + 8y = 0

b. 3x - 8y = 0

c. 8x + 3y = 0

d. 8x = 3y

Solution 10

The Centroid of the triangle is given by

x=-8 and y=3 satisfy(a) 3x + 8y = 0.

Question 11

The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are

(a) (3, 0)

(b) (0, 2)

(c) (2, 3)

(d) (3, 2)

Solution 11

begin mathsize 12px style For space ABCD space to space be space straight a space rectangle
AB space equals space CD space and space BC space equals AD
and space all space angles space must space be space 90 degree. space space space.... open parentheses 1 close parentheses
rightwards double arrow AB space equals space CD space space space and space BC space equals space AD
CB space is space prependicular space distance space of space left parenthesis straight x comma space straight y right parenthesis space from space straight x minus axis
Hence space CB space equals space straight x space space.... open parentheses 2 close parentheses
Similarly comma space CD space is space perpendicular space distance space of space left parenthesis straight x comma space straight y right parenthesis space from space straight y minus axis.
Hence space CD space equals space straight y space space..... open parentheses 3 close parentheses
Also comma space AB space equals 2 space and space AD space equals space 3 space space space space space space space..... open parentheses 4 close parentheses
from space open parentheses 1 close parentheses comma space open parentheses 2 close parentheses comma space open parentheses 3 close parentheses comma space open parentheses 4 close parentheses
straight x space equals space 2 space and space straight y space equals space 3 end style

Question 12

The length of a line segment joining A(2, -3) and B is 10 units. If the abscissa of B is 10 units, then its ordinates can be

(a) 3 or -9

(b) -3 or 9

(c) 6 or 27

(d) -6 or -27

Solution 12

begin mathsize 12px style Let space the space abscissa space be space straight x.
Then space distance space between space points space left parenthesis 2 comma negative 3 right parenthesis space and space left parenthesis 10 comma straight x right parenthesis space is
equals space square root of left parenthesis 2 minus 10 right parenthesis squared plus left parenthesis negative 3 minus straight x right parenthesis squared end root
equals space square root of 8 squared plus left parenthesis straight x plus 3 right parenthesis squared end root
space 10 equals square root of 64 space plus space left parenthesis straight x plus 3 right parenthesis squared end root
rightwards double arrow space 100 space equals space 64 space plus space open parentheses straight x plus 3 close parentheses squared
rightwards double arrow space left parenthesis straight x plus 3 right parenthesis squared space equals space 36
rightwards double arrow straight x space plus space 3 space equals space 6 space or space straight x space plus space 3 space equals space minus 6
rightwards double arrow straight x space equals space 3 space or space straight x equals space minus 9
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 13

The ratio in which the line segment joining P(x1, y1) and Q(x2, y2) is divided by x-axis is

(a) y1 : y2

(b) -y1 : y2

(c) x1 : x2

(d) -x1 : x2

Solution 13

begin mathsize 12px style straight y minus coordinate space of space the space intersection space of space the space straight x minus axis space and space PQ space equals space fraction numerator my subscript 2 space plus space ny subscript 1 over denominator straight m plus straight n end fraction
rightwards double arrow space 0 space equals space fraction numerator my subscript 2 space plus space ny subscript 1 over denominator straight m plus straight n end fraction
rightwards double arrow space my subscript 2 space plus space ny subscript 1 space equals space 0
rightwards double arrow space straight m over straight n equals fraction numerator negative straight y subscript 1 over denominator straight y subscript 2 end fraction
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 14

The ratio in which the line segment joining points A(a1, b1) and B(a2, b2) is divided by y-axis is

(a) -a1 : a2

(b) a1 : a2

(c) b1 : b2

(d) -b1 : b2

Solution 14

begin mathsize 12px style straight x minus coordinate space of space the space intersection space of space the space straight y minus axis space and space AB equals space fraction numerator ma subscript 2 plus na subscript 1 over denominator straight m plus straight n end fraction
space space space space space space space space space space space space space space space space space space space rightwards double arrow space 0 space equals fraction numerator ma subscript 2 plus na subscript 1 over denominator straight m plus straight n end fraction
space space space space space space space space space space space space space space space space space space space rightwards double arrow ma subscript 2 plus na subscript 1 space equals space 0
space space space space space space space space space space space space space space space space space space space rightwards double arrow space straight m over straight n equals fraction numerator negative straight a subscript 1 over denominator straight a subscript 2 end fraction
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Chapter 6 - Co-ordinate Geometry Exercise 6.66

Question 1

 

Solution 1

begin mathsize 12px style straight P space divide s space AB space in space 1 space colon space 2 space ratio
Hence comma
straight P space equals space open parentheses fraction numerator 1 cross times 1 plus 2 cross times 3 over denominator 1 plus 2 end fraction comma space fraction numerator 1 cross times 2 plus 2 cross times open parentheses negative 4 close parentheses over denominator 1 plus 2 end fraction close parentheses
straight P space equals space open parentheses 7 over 3 comma space minus 2 close parentheses
But space straight P space equals space left parenthesis straight a comma space minus 2 right parenthesis
So space on space comparison
straight a equals space 7 over 3
straight Q space divide s space AB space in space 2 space colon space 1
Hence comma space straight Q space equals space open parentheses fraction numerator 2 cross times 1 plus 1 cross times 3 over denominator 2 plus 1 end fraction comma space space fraction numerator 2 cross times 2 minus 1 cross times 4 over denominator 2 plus 1 end fraction close parentheses
space space space space space space space space space space space space space space space space space space equals space open parentheses 5 over 3 comma space 0 close parentheses
Given space straight Q space equals space open parentheses 5 over 3 comma space space straight b close parentheses
Hence comma space straight b space equals space 0
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 2

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are

(a) (-6, 7)

(b) (6, -7)

(c) (6, 7)

(d) -6, -7)

Solution 2

begin mathsize 12px style We space know space that space centre space is space the space mid minus point space of space the space diameter.
Let space the space other space coordinate space of space the space diameter space be space left parenthesis straight x comma straight y right parenthesis.
negative 2 space equals space fraction numerator straight x plus 2 over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space and space space space space space space space space space space space space space space space space space space space space space space space space space space space fraction numerator straight y plus 3 over denominator 2 end fraction equals 5
rightwards double arrow straight x plus 2 space equals space minus 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight y plus 3 equals 10
rightwards double arrow straight x space equals space minus 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight y space equals space 7
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 3

The coordinates of the point P dividing the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1 are

(a) (2, 4)

(b) (3, 5)

(c) (4, 2)

(d) (5, 3)

Solution 3

begin mathsize 12px style Let space straight P left parenthesis straight x comma space straight y right parenthesis space divide space AB space in space 2 space colon space 1
straight x space equals space fraction numerator 2 cross times 4 plus 1 cross times 1 over denominator 2 plus 1 end fraction
space space space equals space 9 over 3
straight x space equals space 3
straight y space equals space fraction numerator 2 cross times 6 plus 1 cross times 3 over denominator 2 plus 1 end fraction
space space space equals space 15 over 3
straight y space equals space 5
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 4

In the figure, the area of ΔABC (in square units) is

(a) 15

(b) 10

(c) 7.5

(d) 2.5

Solution 4

begin mathsize 12px style area space equals space fraction numerator open vertical bar 1 cross times open parentheses 0 minus 0 close parentheses minus 1 open parentheses 0 minus 3 close parentheses plus 4 open parentheses 3 minus 0 close parentheses close vertical bar over denominator 2 end fraction
space space space space space space space space space space equals open vertical bar fraction numerator 3 plus 12 over denominator 2 end fraction close vertical bar
space space space space space space space space space space equals open vertical bar 15 over 2 close vertical bar
space space space space space space space space space space space equals space 7.5
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 5

The point on the x-axis which is equidistant from points (-1, 0) and (5, 0) is

(a) (0, 2)

(b) (2, 0)

(c) (3, 0)

(d) (0, 3)

Solution 5

begin mathsize 12px style Let space point space on space straight x minus axis space be space left parenthesis straight x comma space 0 right parenthesis
straight D subscript 1 space equals space distance space between space left parenthesis negative 1 comma space 0 right parenthesis space and space left parenthesis straight x comma space 0 right parenthesis space equals space square root of open parentheses straight x plus 1 close parentheses squared plus 0 squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open vertical bar straight x plus 1 close vertical bar
straight D subscript 2 space equals space distance space between space left parenthesis 5 comma space 0 right parenthesis space and space left parenthesis straight x comma space 0 right parenthesis space equals space square root of open parentheses straight x minus 5 close parentheses squared plus 0 squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open vertical bar straight x minus 5 close vertical bar
rightwards double arrow space straight D subscript 1 space equals space straight D subscript 2
rightwards double arrow open vertical bar straight x plus 1 close vertical bar equals open vertical bar straight x minus 5 close vertical bar
rightwards double arrow straight x plus 1 equals negative open parentheses straight x minus 5 close parentheses
rightwards double arrow straight x plus 1 space equals space minus straight x plus 5
rightwards double arrow 2 straight x space equals space 4
rightwards double arrow straight x space equals space 2
left parenthesis 2 comma space 0 right parenthesis space is space co minus ordinate
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 6

begin mathsize 12px style If space straight A left parenthesis 4 comma 9 right parenthesis comma space straight B left parenthesis 2 comma 3 right parenthesis space and space straight C left parenthesis 6 comma 5 right parenthesis space are space the space vertices space of space triangle ABC comma space then space the space length space of space median space through space straight C space is
left parenthesis straight a right parenthesis space 5 space units
left parenthesis straight b right parenthesis space square root of 10 space units
left parenthesis straight c right parenthesis space 25 space units
left parenthesis straight d right parenthesis space 10 space units end style

Solution 6

begin mathsize 12px style straight M space is space mid minus point space of space AB
straight M space equals space open parentheses fraction numerator 4 plus 2 over denominator 2 end fraction comma space fraction numerator 9 plus 3 over denominator 2 end fraction close parentheses
space space space space equals space open parentheses 3 comma space 6 close parentheses
straight M space equals space square root of open parentheses 6 minus 3 close parentheses squared plus open parentheses 5 minus 6 close parentheses squared end root
space space space space space equals space square root of 3 squared plus 1 squared end root
space space space space space space equals square root of 10
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 7

If P(2, 4), Q(0,3), R (3, 6) and S (5, y) are the vertices of a parallelogram PQRS, then the value of y is

(a) 7

(b) 5

(c) -7

(d) -8

Solution 7