R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 6 - Co-ordinate Geometry

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Chapter 16 - Co-ordinate Geometry Ex. 16D

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Solution 10

Distance between the points

         

Solution 11

  

Solution 12

The points A(4,3) and B(x, 5) lie on the circle with center O(2,3)

OA and OB are radius of the circle.

Solution 13

The point P(x, y) is equidistant from the point A(7, 1) and B(3, 5)

Solution 14

The vertices of ABC are (a, b), (b, c) and (c, a)

Centroid is

But centroid is (0, 0)

a + b + c = 0

Solution 15

The vertices of ABC are A(2, 2), B(-4, -4) and C(5, -8)

Centroid of ABC is given by

Solution 16

Let the point C(4, 5) divides the join of A(2, 3) and B(7, 8) in the ratio k : 1

The point C is

But C is (4, 5)

Thus, C divides AB in the ratio 2 : 3

Solution 17

The points A(2, 3), B(4, k) and C(6, -3) are collinear if area of ABC is zero

But area of ABC = 0,

k = 0

Chapter 16 - Co-ordinate Geometry Ex. 6C

Solution 9

  

  

 

Solution 1

(i)Let A(1, 2), B(-2, 3) and C(-3, -4) be the vertices ofthe given ABC, then

   

(ii)The coordinates of vertices of ABC are A(-5, 7), B(-4, -5) and C(4, 5)

Here,

(iii)The coordinates of ABC are A(3, 8), B(-4, 2) and C(5, -1)

   

 

(iv)Let P(10, -6), Q(2, 5) and R(-1, 3) be the vertices of the given PQR. Then,

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Solution 10(i)

  

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Solution 13(i)

  

Solution 13(ii)

  

Solution 13(iii)

  

Solution 13(iv)

  

Solution 14

  

Solution 15

The given points are A(-3, 12), B(7, 6) and C(x, 9)

Solution 16

Let P(1, 4), Q(3, y) and R(-3, 16)

Solution 17

  

Solution 18

  

Solution 19

Vertices of ABC are A(2, 1), B(x, y) and C(7, 5)

The points A, B and C are collinear

area of ABC =0

Or 4x - 5y - 3 = 0

Solution 20

  

Solution 21

The vertices of ABC are (a, 0), (0, b), C(1, 1)

 

The points A, B, C are collinear

Area of ABC = 0

ab - a - b = 0 a + b = ab

Dividing by ab

Solution 22

  

Solution 23

  

  

Chapter 16 - Co-ordinate Geometry Ex. 6A

Solution 12

Let A(6, -1) and B(2,3) be the given point and P(x,y) be the required point, we get

Solution 1

(i)The given points are A(9,3) and B(15,11)

 

 

(ii)The given points are A(7,4) and B(-5,1)

 

 

(iii)The given points are A(-6, -4) and B(9,-12)

 

 

(iv)The given points are A(1, -3) and B(4, -6)

 

 

(v)The given points are P(a + b, a - b) and Q(a - b, a + b)

(vi)The given points are P(a sin a, a cos a) and Q(a cos a, - a sina)

begin mathsize 12px style left parenthesis straight X subscript 1 equals asin space straight a comma straight y subscript 1 equals straight a space cos space straight a right parenthesis space and space left parenthesis straight x subscript 2 equals acos space straight a comma space straight y subscript 2 equals negative aain space straight a right parenthesis
PQ equals square root of left parenthesis straight x subscript 2 minus straight x subscript 1 right parenthesis squared plus left parenthesis straight y subscript 2 minus straight y subscript 1 right parenthesis squared end root
space space space space space space equals square root of left parenthesis straight a space cos space straight a space minus space straight a space sin space straight a right parenthesis squared plus left parenthesis negative straight a space cos space straight a right parenthesis to the power of 2 end exponent end root
equals square root of straight a squared cos squared straight a plus straight a squared sin squared straight a space minus 2 straight a squared cos space straight a space sin space straight a plus straight a squared cos squared straight a plus straight a squared sin squared straight a plus 2 straight a squared cos space straight a space sin space straight a end root
space space space space space equals square root of straight a squared cos squared straight a plus straight a squared sin squared straight a plus straight a squared cos squared end root straight a plus straight a squared sin squared straight a
space space space space space equals square root of straight a squared left parenthesis cos squared straight a plus sin squared straight a right parenthesis plus straight a squared left parenthesis cos squared straight a plus sin squared straight a right parenthesis end root
space space space space space equals space square root of straight a squared plus straight a squared end root equals square root of 2 straight a squared end root equals square root of 2 straight a space units end style

Solution 2

(i)The given point is A(5, -12) and let O(0,0) be the origin

(ii)The given point is B(-5, 5) and let O(0,0) be the origin


(iii)The given point is C(-4, -6) and let O(0,0) be the origin

Solution 3

The given points are A(a, -1) and B(5,3)

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Solution 8

Let any point P on x - axis is (x,0) which is equidistant from A(-2, 5) and B(-2, 9)

This is not admissible

Hence, there is no point on x - axis which is equidistant from A(-2, 5) and B(-2, 9)

Solution 9

Let A(11, -8) be the given point and let P(x,0) be the required point on x - axis

Then,

Hence, the required points are (17,0) and (5,0)

Solution 10

  

Solution 11

  

Solution 13

Let the required points be P(x,y), then

PA = PB = PC. The points A, B, C are (5,3), (5, -5) and (1, -5) respectively

Hence, the point P is (3, -1)

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Solution 16

6A  

Solution 17

  

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Solution 18(ii)

  

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Solution 24

Let A(-5,6), B(3,0) and C(9,8) be the given points. Then

Solution 25

are the given points

Hence, DABC is equilateral and each of its sides being

Solution 26

 

(i)The angular points of quadrilateral ABCD are A(3,2), B(0,5), C(-3,2) and D(0,-1)

 

 

     Thus, all sides of quad. ABCD are equal and diagonals are also equal

 

      Quad. ABCD is a square

(ii)Let A(6,2), B(2,1), C(1,5) and D(5,6) be the angular points of quad. ABCD. Join AC and BD


Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.

Hence, quad ABCD is a square.


(iii)Let P(0, -2), Q(3,1), R(0,4) and S(-3,1) be the angular points of quad. ABCD


Join PR and QSD


Thus, PQRS is a quadrilateral in which all sides are equal and the diagonals are equal

Hence, quad. PQRS is a square

Solution 27

Let A(-3,2), B(-5, -5), C(2, -3) and D(4,4) be the angular point of quad ABCD. Join AC and BD.

 

 

Thus, ABCD is a quadrilateral having all sides equal but diagonals are unequal.

Hence, ABCD is a rhombus

Solution 28

  

  

Solution 29

  

  

Solution 30

Let A(2,1), B(5,2), C(6,4) and D(3,3) are the angular points of a parallelogram ABCD. Then

Diagonal AC  Diagonal BD

Thus ABCD is not a rectangle but it is a parallelogram because its opposite sides are equal and diagonals are not equal

Solution 31

  

Solution 32

 

(i) Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) are the vertices of quad. ABCD. Then

 

           

 

 

 

Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal

 

Hence, quad. ABCD is a rectangle.

(ii)Let A(2, -2), B(14, 10), C(11, 13) and D(-1, 1) be the angular points of quad. ABCD, then

Thus, ABCD is a quadrilateral whose opposite sides are equal and diagonals are equal.

Hence, quad. ABCD is rectangle.

(iii)Let A(0, -4), B(6,2), C(3,5) and D(-3,-1) are the vertices of quad. ABCD. Then

 

 

 

 

 

  Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal

 

Hence, quad. ABCD is a rectangle


Chapter 16 - Co-ordinate Geometry MCQ

Solution 1

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Correct option: (d)

  

Solution 7

Correct option: (b)

  

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Solution 12

Correct option: (a)

  

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Solution 33

  

Solution 34

Chapter 6 - Co-ordinate Geometry Ex. 6B

Solution 1

The end points of AB are A(-1,7) and B(4, -3)

Let the required point be P(x, y)

By section formula, we have

Hence the required point is P(1, 3)

Solution 2

The end points of PQ are P(-5, 11) and Q(4, -7_

By section formula, we have

Hence the required point is (2, -3)

Solution 3

  

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Solution 6

Points P, Q, R divide the line segment joining the points A(1,6) and B(5, -2) into four equal parts

Point P divide AB in the ratio 1 : 3 where A(1, 6), B(5, -2)

Therefore,  the point P is

 

Also, R is the midpoint of the line segment joining Q(3, 2) and B(5, -2)

Solution 7

Point P divides the join of A(3, -4) and B(1,2) in the ratio 1 : 2.

Coordinates of P are:

Solution 8

(i)The coordinates of mid - points of the line segment joining A(3, 0) and B(-5, 4) are

(ii)Let M(x, y) be the mid - point of AB, where A is (-11, -8) and B is (8, -2). Then,

Solution 9

The midpoint of line segment joining the points A(6, -5) and B(-2, 11) is

Also, given the midpoint of AB is (2, p)

p = 3

Solution 10

C(1, 2a + 1) is the midpoint of A(2a, 4) and B(-2, 3b)

Solution 11

Let A(-2, 9) and B(6, 3) be the two points of the given diameter AB and let C(a, b) be the center of the circle

Then, clearly C is the midpoint of AB

By the midpoint formula of the co-ordinates,

Hence, the required point C(2, 6)

Solution 12

A, B are the end points of a diameter. Let the coordinates of A be (x, y)

The point B is (1, 4)

The center C(2, -3) is the midpoint of AB

The point A is (3, -10)

Solution 13

Let P divided the join of A(8, 2), B(-6, 9) in the ratio k : 1

By section formula, the coordinates of p are

Hence, the required ratio of which is (3 : 4)

Solution 14

  

Solution 15

Let P divided the join of line segment A(-4, 3) and B(2, 8) in the ratio k : 1

the point P is

Solution 16

Let P is dividing the given segment joining A(-5, -4) and B(-2, 3) in the ratio r : 1

Coordinates of point P


 

Solution 17

Let the x- axis cut the join of A(2, -3) and B(5, 6) in the ratio k : 1 at the point P

Then, by the section formula, the coordinates of P are

But P lies on the x axis so, its ordinate must be 0

So the required ratio is 1 : 2

Thus the x - axis divides AB in the ratio 1 : 2

Putting we get the point P as

Thus, P is (3, 0) and k = 1 : 2

Solution 18

Let the y - axis cut the join A(-2, -3) and B(3, 7) at the point P in the ratio k : 1

Then, by section formula, the co-ordinates of P are

But P lies on the y-axis so, its abscissa is 0

So the required ratio is which is 2 : 3

Putting we get the point P as

i.e., P(0, 1)

Hence the point of intersection of AB and the y - axis is P(0, 1) and P divides AB in the ratio 2 : 3

Solution 19

Let the line segment joining A(3, -1) and B(8, 9) is divided byx - y - 2 = 0 in ratio k : 1 at p

Coordinates of P are

Thus the line x - y - 2 = 0 dividesAB in the ratio 2 : 3

Solution 20

Let D, E, F be the midpoint of the side BC, CA and AB respectively in ABC


Then, by the midpoint formula, we have

Hence the lengths of medians AD, BE and CF are given by

Solution 21

Here

Let G(x, y) be the centroid of ABC, then

Hence the centroid of ABC is G(4, 0)

Solution 22

Two vertices of ABC are A(1, -6) and B(-5, 2) let the third vertex be C(a, b)

Then, the co-ordinates of its centroid are

But given that the centroid is G(-2, 1)

Hence, the third vertex C of ABC is (-2, 7)

Solution 23

Two vertices of ABC are B(-3, 1) and C(0, -2) and third vertex be A(a, b)

Then the coordinates of its centroid are

Hence the third vertices A of ABC is A(3, 1)

Solution 24

Let A(3,1), B(0, -2), C(1, 1) and D(4, 4) be the vertices of quadrilateral

Join AC, BD. AC and BD, intersect other at the point O.

We know that the diagonals of a parallelogram bisect each other

Therefore, O is midpoint of AC as well as that of BD

Now midpoint of AC is

And midpoint of BD is

Mid point of AC is the same as midpoint of BD

Hence, A, B, C, D are the vertices of a parallelogram ABCD

Solution 25

Let P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS.

Join the diagonals PR and SQ.

They intersect each other at the point O. We know that the diagonals of a parallelogram bisect each other.

Therefore, O is the midpoint of PR as well as that of SQ

Now, midpoint of PR is

And midpoint of SQ is

Hence the required values are a = 4 and b = 3

Solution 26

Let A(1, -2), B(3, 6) and C(5, 10) are the given vertices of the parallelogram ABCD

Let D(a, b) be its fourth vertex. Join AC and BD.

Let AC and BD intersect at the point O.

We know that the diagonals of a parallelogram bisect each other.

So, O is the midpoint AC as well as that of BD

Midpoint of AC is

Midpoint of BD is

Hence the fourth vertices is D(3, 2)

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Solution 34