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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line

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Selina Textbook Solutions Chapter 14 - Equation of a Line

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 14 - Equation of a Line.

All Selina textbook questions of Chapter 14 - Equation of a Line solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam. 

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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(A)

Solution 1

The given line is x - 2y + 5 = 0.


(i) Substituting x = 1 and y = 3 in the given equation, we have:

L.H.S. = 1 - 2 3 + 5 = 1 - 6 + 5 = 6 - 6 = 0 = R.H.S.

Thus, the point (1, 3) lies on the given line.


 

(ii) Substituting x = 0 and y = 5 in the given equation, we have:

L.H.S. = 0 - 2 5 + 5 = -10 + 5 = -5 R.H.S.

Thus, the point (0, 5) does not lie on the given line.


 

(iii) Substituting x = -5 and y = 0 in the given equation, we have:

L.H.S. = -5 - 2 0 + 5 = -5 - 0 + 5 = 5 - 5 = 0 = R.H.S.

Thus, the point (-5, 0) lie on the given line.


 

(iv) Substituting x = 5 and y = 5 in the given equation, we have:

L.H.S. = 5 - 2 5 + 5 = 5 - 10 + 5 = 10 - 10 = 0 = R.H.S.

Thus, the point (5, 5) lies on the given line.


 

(v) Substituting x = 2 and y = -1.5 in the given equation, we have:

L.H.S. = 2 - 2 (-1.5) + 5 = 2 + 3 + 5 = 10 R.H.S.

Thus, the point (2, -1.5) does not lie on the given line.


 

(vi) Substituting x = -2 and y = -1.5 in the given equation, we have:

L.H.S. = -2 - 2 (-1.5) + 5 = -2 + 3 + 5 = 6 R.H.S.

Thus, the point (-2, -1.5) does not lie on the given line.

Solution 2

(i) The given line is

Substituting x = 2 and y = 3 in the given equation,

Thus, the given statement is false.

(ii) The given line is

Substituting x = 4 and y = -6 in the given equation,

Thus, the given statement is true.

(iii) L.H.S = y - 7 = 7 - 7 = 0 = R.H.S.

Thus, the point (8, 7) lies on the line y - 7 = 0.

The given statement is true.

(iv) L.H.S. = x + 3 = -3 + 3 = 0 = R.H.S

Thus, the point (-3, 0) lies on the line x + 3 = 0.

The given statement is true.

(v) The point (2, a) lies on the line 2x - y = 3.

2(2) - a = 3

4 - a = 3

a = 4 - 3 = 1

Thus, the given statement is false.

Solution 3

Given, the line given by the equation passes through the point (k, 6).

Substituting x = k and y = 6 in the given equation, we have:

Solution 4

The given equation of the line is 9x + 4y = 3.

Put x = 3 and y = -k, we have:

9(3) + 4(-k) = 3

27 - 4k = 3

4k = 27 - 3 = 24

k = 6

Solution 5

The equation of the given line is

Putting x = m, y = 2m - 1, we have:


Solution 6

The given line will bisect the join of A (5, -2) and B (-1, 2), if the co-ordinates of the mid-point of AB satisfy the equation of the line.

The co-ordinates of the mid-point of AB are

Substituting x = 2 and y = 0 in the given equation, we have:

L.H.S. = 3x - 5y = 3(2) - 5(0) = 6 - 0 = 6 = R.H.S.

Hence, the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2).

Solution 7

(i) The given line bisects the join of A (a, 3) and B (2, -5), so the co-ordinates of the mid-point of AB will satisfy the equation of the line.

The co-ordinates of the mid-point of AB are

Substituting x = and y = -1 in the given equation, we have:

(ii) The given line bisects the join of A (8, -1) and B (0, k), so the co-ordinates of the mid-point of AB will satisfy the equation of the line.

The co-ordinates of the mid-point of AB are

Substituting x = 4 and y = in the given equation, we have:

Solution 8

(i) Given, the point (-3, 2) lies on the line ax + 3y + 6 = 0.

Substituting x = -3 and y = 2 in the given equation, we have:

a(-3) + 3(2) + 6 = 0

-3a + 12 = 0

3a = 12

a = 4

(ii) Given, the line y = mx + 8 contains the point (-4, 4).

Substituting x = -4 and y = 4 in the given equation, we have:

4 = -4m + 8

4m = 4

m = 1

Solution 9

Given, the point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3.

Co-ordinates of the point P are

Substituting x = 0 and y = 3 in the given equation, we have:

L.H.S. = 0 - 5(3) + 15 = -15 + 15 = 0 = R.H.S.

Hence, the point P lies on the line x - 5y + 15 = 0.

Solution 10

Given, the line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1: 2.

Co-ordinates of the point Q are

Substituting x = 4 and y = -2 in the given equation, we have:

L.H.S. = x - 2y = 4 - 2(-2) = 4 + 4 = 8 R.H.S.

Hence, the given line does not contain point Q.

Solution 11


Consider the given equations:

4x + 3y = 1 ....(1)

3x - y + 9 = 0 ....(2)

Multiplying (2) with 3, we have:

9x - 3y = -27 ....(3)

Adding (1) and (3), we get,

13x = -26

x = -2

 

From (2), y = 3x + 9 = -6 + 9 = 3

 

Thus, the point of intersection of the given lines (1) and (2) is (-2, 3).

 

The point (-2, 3) lies on the line (2k - 1)x - 2y = 4.

(2k - 1)(-2) - 2(3) = 4

-4k + 2 - 6 = 4

-4k = 8

k = -2

Solution 12

We know that two or more lines are said to be concurrent if they intersect at a single point.

 

We first find the point of intersection of the first two lines.

2x + 5y = 1 ....(1)

x - 3y = 6 ....(2)

 

Multiplying (2) by 2, we get,

2x - 6y = 12 ....(3)

 

Subtracting (3) from (1), we get,

11y = -11

y = -1

 

From (2), x = 6 + 3y = 6 - 3 = 3

 

So, the point of intersection of the first two lines is (3, -1).

If this point lie on the third line, i.e., x + 5y + 2 = 0, then the given lines will be concurrent.

 

Substituting x = 3 and y = -1, we have:

L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 5 - 5 = 0 = R.H.S.

 

Thus, (3, -1) also lie on the third line.

Hence, the given lines are concurrent.

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(E)

Solution 1

Using section formula, the co-ordinates of the point P are

 

3x + 5y = 7

Slope of this line =

As the required line is parallel to the line 3x + 5y = 7,

Slope of the required line = Slope of the given line =

Thus, the equation of the required line is

y - y1 = m(x - x1)

y + 3 = (x - 11)

5y + 15 = -3x + 33

3x + 5y = 18

Solution 2

Using section formula, the co-ordinates of the point P are

The equation of the given line is

5x - 3y + 4 = 0

Slope of this line =

Since, the required line is perpendicular to the given line,

Slope of the required line =

Thus, the equation of the required line is

y - y1 = m(x - x1)

Solution 3

Point P lies on y-axis, so putting x = 0 in the equation 5x + 3y + 15 = 0, we get, y = -5

Thus, the co-ordinates of the point P are (0, -5).

x - 3y + 4 = 0

Slope of this line =

The required equation is perpendicular to given equation x - 3y + 4 = 0.

Slope of the required line =

(x1, y1) = (0, -5)

Thus, the required equation of the line is

y - y1 = m(x - x1)

y + 5 = -3(x - 0)

3x + y + 5 = 0

Solution 4

kx - 5y + 4 = 0

Slope of this line = m1 =

5x - 2y + 5 = 0

Slope of this line = m2 =

Since, the lines are perpendicular, m1m2 = -1

Solution 5

(i) Slope of PQ =

Equation of the line PQ is given by

y - y1 = m(x - x1)

y - 4 = -1(x + 1)

y - 4 = -x - 1

x + y = 3

(ii) For point A (on x-axis), y = 0.

Putting y = 0 in the equation of PQ, we get,

x = 3

Thus, the co-ordinates of point A are (3, 0).

For point B (on y-axis), x = 0.

Putting x = 0 in the equation of PQ, we get,

y = 3

Thus, the co-ordinates of point B are (0, 3).

(iii) M is the mid-point of AB.

So, the co-ordinates of point M are

Solution 6

A = (1, 5) and C = (-3, -1)

We know that in a rhombus, diagonals bisect each other at right angle.

Let O be the point of intersection of the diagonals AC and BD.

Co-ordinates of O are

Slope of AC =

For line AC:

Slope = m = , (x1, y1) = (1, 5)

Equation of the line AC is

y - y1 = m(x - x1)

y - 5 = (x - 1)

2y - 10 = 3x - 3

3x - 2y + 7 = 0

For line BD:

Slope = m = , (x1, y1) = (-1, 2)

Equation of the line BD is

y - y1 = m(x - x1)

y - 2 = (x + 1)

3y - 6 = -2x - 2

2x + 3y = 4

Solution 7

 

U sin g space d i s tan c e space f o r m u l a comma space w e space h a v e :
A B equals square root of open parentheses 6 minus 3 close parentheses squared plus open parentheses minus 2 minus 2 close parentheses squared end root equals square root of 9 plus 16 end root equals 5
B C equals square root of open parentheses 2 minus 6 close parentheses squared plus open parentheses minus 5 plus 2 close parentheses squared end root equals square root of 16 plus 9 end root equals 5
T h u s comma space A C equals B C
A l s o comma space S l o p e space o f space A B equals fraction numerator minus 2 minus 2 over denominator 6 minus 3 end fraction equals fraction numerator minus 4 over denominator 3 end fraction
S l o p e space o f space B C equals fraction numerator minus 5 plus 2 over denominator 2 minus 6 end fraction equals fraction numerator minus 3 over denominator minus 4 end fraction equals 3 over 4
S l o p e space o f space A B cross times S l o p e space o f space B C equals minus 1
T h u s comma space A B perpendicular B C
H e n c e comma space A comma space B comma space C space c a n space b e space t h e space v e r t i c e s space o f space a space s q u a r e..
open parentheses i close parentheses space space S l o p e space o f space A B equals fraction numerator minus 2 minus 2 over denominator 6 minus 3 end fraction equals equals S l o p e space o f space C D
E q u a t i o n space o f space t h e space l i n e space C D space i s
y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
rightwards double arrow y plus 5 equals fraction numerator minus 4 over denominator 3 end fraction open parentheses x minus 2 close parentheses
rightwards double arrow 3 y plus 15 equals minus 4 x plus 8
rightwards double arrow 4 x plus 3 y equals minus 7.... left parenthesis 1 right parenthesis
S l o p e space o f space B C equals fraction numerator minus 5 plus 2 over denominator 2 minus 6 end fraction equals fraction numerator minus 3 over denominator minus 4 end fraction equals 3 over 4 equals S l o p e space o f space A D
E q u a t i o n space o f space t h e space l i n e space A D space i s
y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
rightwards double arrow y minus 2 equals 3 over 4 open parentheses x minus 3 close parentheses
rightwards double arrow 4 y minus 8 equals 3 x minus 9
rightwards double arrow 3 x minus 4 y equals 1... left parenthesis 2 right parenthesis
N o w comma space D space i s space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space C D space a n d space A D.
left parenthesis 1 right parenthesis rightwards double arrow 16 x plus 12 y equals minus 28
open parentheses 2 close parentheses rightwards double arrow 9 x minus 12 y equals 3
A d d i n g space t h e space a b o v e space t w o space e q u a t i o n s space w e space g e t comma
25 x equals minus 25
rightwards double arrow x equals minus 1
S o comma space 4 y equals 3 x minus 1 equals minus 3 minus 1 equals minus 4
rightwards double arrow y equals minus 1
T h u s comma space t h e space c o minus o r d i n a t e s space o f space p o i n t space D space a r e space left parenthesis minus 1 comma space minus 1 right parenthesis.

left parenthesis i i right parenthesis
T h e space e q u a t i o n space o f space l i n e space A D space i s space f o u n d space i n space p a r t space left parenthesis i right parenthesis
I t space i s space 3 x minus 4 y equals 1 space o r space 4 y equals 3 x minus 1.
S l o p e space o f space B D equals fraction numerator minus 1 plus 2 over denominator minus 1 minus 6 end fraction equals fraction numerator 1 over denominator minus 7 end fraction equals fraction numerator minus 1 over denominator 7 end fraction
T h e space e q u a t i o n space o f space d i a g o n a l space B D space i s
y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
rightwards double arrow y plus 1 equals fraction numerator minus 1 over denominator 7 end fraction open parentheses x plus 1 close parentheses
rightwards double arrow 7 y plus 7 equals minus x minus 1
rightwards double arrow x plus 7 y plus 8 equals 0

 

Solution 8

The given line is

x = 3y + 2 ...(1)

3y = x - 2

y equals 1 third x minus 2 over 3

Slope of this line is 1 third.

The required line intersects the given line at right angle.

Slope of the required line =fraction numerator minus 1 over denominator begin display style 1 third end style end fraction equals minus 3 

The required line passes through (0, 0) = (x1, y1)

The equation of the required line is

y - y1 = m(x - x1)

y - 0 = -3(x - 0)

3x + y = 0 ...(2)

 

Point X is the intersection of the lines (1) and (2).

Using (1) in (2), we get,

9y + 6 + y = 0

y equals fraction numerator minus 6 over denominator 10 end fraction equals fraction numerator minus 3 over denominator 5 end fraction

Thus, the co-ordinates of the point X are open parentheses 1 fifth comma fraction numerator minus 3 over denominator 5 end fraction close parentheses.

Solution 9

Let the line intersect the x-axis at point A (x, 0) and y-axis at point B (0, y).

Since, P is the mid-point of AB, we have:

Thus, A = (6, 0) and B = (0, 4)

Slope of line AB =

Let (x1, y1) = (6, 0)

The required equation of the line AB is given by

y - y1 = m(x - x1)

y - 0 = (x - 6)

3y = -2x + 12

2x + 3y = 12

Solution 10

7x + 6y = 71 28x + 24 = 284 ...(1)

5x - 8y = -23 15x - 24y = -69 ...(2)

Adding (1) and (2), we get,

43x = 215

x = 5

 

From (2), 8y = 5x + 23 = 25 + 23 = 48 y = 6

 

Thus, the required line passes through the point (5, 6).

 

4x - 2y = 1

2y = 4x - 1

y = 2x -

Slope of this line = 2

Slope of the required line =

The required equation of the line is

y - y1 = m(x - x1)

y - 6 = (x - 5)

2y - 12 = -x + 5

x + 2y = 17

Solution 11

The given line is

Slope of this line =

 

Slope of the required line =

 

Let the required line passes through the point P (0, y).

Putting x = 0 in the equation , we get,

Thus, P = (0, -b) = (x1, y1)

 

The equation of the required line is

y - y1 = m(x - x1)

y + b = (x - 0)

by + b2 = -ax

ax + by + b2 = 0

Solution 12

(i) Let the median through O meets AB at D. So, D is the mid-point of AB.

Co-ordinates of point D are

Slope of OD =

(x1, y1) = (0, 0)

The equation of the median OD is

y - y1 = m(x - x1)

y - 0 = -1(x - 0)

x + y = 0

(ii) The altitude through vertex B is perpendicular to OA.

Slope of OA =

Slope of the required altitude =

The equation of the required altitude through B is

y - y1 = m(x - x1)

y + 3 = (x + 5)

5y + 15 = -3x - 15

3x + 5y + 30 = 0

Solution 13

Let A = (-2, 3) and B = (4, 1)

Slope of AB = m1 =

Equation of line AB is

y - y1 = m1(x - x1)

y - 3 = (x + 2)

3y - 9 = -x - 2

x + 3y = 7 ...(1)

 

Slope of the given line 3x = y + 1 is 3 = m2.

Hence, the line through points A and B is perpendicular to the given line.

 

Given line is 3x = y +1 ...(2)

 

Solving (1) and (2), we get,

x = 1 and y = 2

 

So, the two lines intersect at point P = (1, 2).

 

The co-ordinates of the mid-point of AB are

Hence, the line 3x = y + 1 bisects the line segment joining the points A and B.

Solution 14

x cos  + y sin  = 2

Slope of this line =

Slope of a line which is parallel to this given line =

Let (4, 3) = (x1, y1)

Thus, the equation of the required line is given by:

y - y1 = m1(x - x1)

y - 3 = (x - 4)

Solution 15

(k - 2)x + (k + 3)y - 5 = 0 ....(1)

(k + 3)y = -(k - 2)x + 5

y =

Slope of this line = m1 =

(i) 2x - y + 7 = 0

y = 2x + 7 = 0

Slope of this line = m2 = 2

 

Line (1) is perpendicular to 2x - y + 7 = 0

 

(ii) Line (1) is parallel to 2x - y + 7 = 0

Solution 16

Slope of BC =

Equation of the line BC is given by

y - y1 = m1(x - x1)

y + 2 = (x + 1)

4y + 8 = 3x + 3

3x - 4y = 5....(1)

 

(i) Slope of line perpendicular to BC =

Required equation of the line through A (0, 5) and perpendicular to BC is

y - y1 = m1(x - x1)

y - 5 = (x - 0)

3y - 15 = -4x

4x + 3y = 15 ....(2)

 

(ii) The required point will be the point of intersection of lines (1) and (2).

 

(1) 9x - 12y = 15

(2) 16x + 12y = 60

 

Adding the above two equations, we get,

25x = 75

x = 3

 

So, 4y = 3x - 5 = 9 - 5 = 4

y = 1

 

Thus, the co-ordinates of the required point is (3, 1).

Solution 17

(i) A = (2, 3), B = (-1, 2), C = (3, 0)

(ii) Slope of BC =

Slope of required line which is parallel to BC = Slope of BC =

(x1, y1) = (2, 3)

The required equation of the line through A and parallel to BC is given by:

y - y1 = m1(x - x1)

y - 3 = (x - 2)

2y - 6 = -x + 2

x + 2y = 8

Solution 18

The median (say RX) through R will bisect the line PQ.

The co-ordinates of point X are

Slope of RX =

(x1, y1) = (-2, -1)

The required equation of the median RX is given by:

y - y1 = m1(x - x1)

y + 1 = (x + 2)

7y + 7 = 2x + 4

7y = 2x - 3

Solution 19

P is the mid-point of AB. So, the co-ordinate of point P are

Q is the mid-point of AC. So, the co-ordinate of point Q are

Slope of PQ =

Slope of BC =

Since, slope of PQ = Slope of BC,

PQ || BC

Also, we have:

Slope of PB =

Slope of QC =

Thus, PB is not parallel to QC.

Hence, PBCQ is a trapezium.

Solution 20

(i) Let the co-ordinates of point A (lying on x-axis) be (x, 0) and the co-ordinates of point B (lying y-axis) be (0, y).

Given, P = (-4, -2) and AP: PB = 1:2

Using section formula, we have:

Thus, the co-ordinates of A and B are (-6, 0) and (0, -6).

(ii) Slope of AB =

Slope of the required line perpendicular to AB =

(x1, y1) = (-4, -2)

Required equation of the line passing through P and perpendicular to AB is given by

y - y1 = m(x - x1)

y + 2 = 1(x + 4)

y + 2 = x + 4

y = x + 2

Solution 21

The required line intersects x-axis at point A (-2, 0).

Also, y-intercept = 3

So, the line also passes through B (0, 3).

Slope of line AB = = m

(x1, y1) = (-2, 0)

Required equation of the line AB is given by

y - y1 = m(x - x1)

y - 0 = (x + 2)

2y = 3x + 6

Solution 22

The required line passes through A (2, 3).

Also, x-intercept = 4

So, the required line passes through B (4, 0).

Slope of AB =

(x1, y1) = (4, 0)

Required equation of the line AB is given by

y - y1 = m(x - x1)

y - 0 = (x - 4)

2y = -3x + 12

3x + 2y = 12

Solution 23

Equation of the line AB is y = x + 1

Slope of AB = 1

Inclination of line AB = (Since, tan 45o = 1)

Equation of line CD is y = x - 1

Slope of CD =

Inclination of line CD = 60o (Since, tan 60o =)

Using angle sum property in PQR,

Solution 24

Given, P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2: 3.

Co-ordinates of point P are

Slope of the required line = m =

The required equation of the line is given by

y - y1 = m(x - x1)

y - 2 = (x - 0)

2y - 4 = 3x

2y = 3x + 4

Solution 25

Let A = (6, 4) and B = (7, -5)

Slope of the line AB =

(x1, y1) = (6, 4)

 

The equation of the line AB is given by

y - y1 = m(x - x1)

y - 4 = -9(x - 6)

y - 4 = -9x + 54

9x + y = 58 ...(1)

 

Now, given that the ordinate of the required point is -23.

Putting y = -23 in (1), we get,

9x - 23 = 58

9x = 81

x = 9

 

Thus, the co-ordinates of the required point is (9, -23).

Solution 26

Given points are A(7, -3) and B(1, 9).

(i) Slope of AB =

(ii) Slope of perpendicular bisector = =

Mid-point of AB = =(4, 3)

Equation of perpendicular bisector is:

y - 3 = (x - 4)

2y - 6 = x - 4

x - 2y + 2 = 0

(iii) Point (-2, p) lies on x - 2y + 2 = 0.

-2 - 2p + 2 = 0

2p = 0

p = 0

Solution 27

(i) Let the co-ordinates be A(x, 0) and B(0, y).

Mid-point of A and B is given by

(ii) Slope of line AB, m =

(iii) Equation of line AB, using A(4, 0)

2y = 3x - 12

Solution 28

3x + 4y - 7 = 0 ...(1)

4y = -3x + 7

y =

(i) Slope of the line = m =

(ii) Slope of the line perpendicular to the given line =

Solving the equations x - y + 2 = 0 and 3x + y - 10 = 0, we get x = 2 and y = 4.

So, the point of intersection of the two given lines is (2, 4).

Given that a line with slope passes through point (2, 4).

Thus, the required equation of the line is

y - 4 = (x - 2)

3y - 12 = 4x - 8

4x - 3y + 4 = 0

Solution 29

 

In parallelogram ABCD, A(x, y), B(5, 8), C(4, 7) and D(2, -4).

The diagonals of the parallelogram bisect each other.

O is the point of intersection of AC and BD

Since O is the midpoint of BD, its coordinates will be

 

(i)

Since O is the midpoint of AC also,

(ii)

Solution 30

(i)

 

(ii)

 

 

(iii)

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

  

 

 

Solution 41

i. Since A lies on the X-axis, let the co-ordinates of A be (x, 0).

Since B lies on the Y-axis, let the co-ordinates of B be (0, y).

Let m = 1 and n = 2

Using Section formula,

x = 6 and y = -3

So, the co-ordinates of A are (6, 0) and that of B are (0, -3). 

 

Slope of line perpendicular to AB = m = -2

P = (4, -1)

Thus, the required equation is

y - y1 = m(x - x1)

y - (-1) = -2(x - 4)

y + 1 = -2x + 8

2x + y = 7 

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(B)

Solution 1

(i) Slope = tan 0o = 0

(ii) Slope = tan 30o =

(iii) Slope = tan 72o 30' = 3.1716

(iv) Slope = tan 46o = 1.0355

Solution 2

(i) Slope = tan = 0

= 0o

(ii) Slope = tan =

= 60o

(iii) Slope = tan = 0.7646

= 37o 24'

(iv) Slope = tan = 1.0875

= 47o 24'

Solution 3

We know:

Slope =

(i) Slope =

(ii) Slope =

(iii) Slope =

Solution 4

(i) Slope of AB =

Slope of the line parallel to AB = Slope of AB = 1

(ii) Slope of AB =

Slope of the line parallel to AB = Slope of AB = -4

Solution 5

(i) Slope of AB =

Slope of the line perpendicular to AB =

(ii) Slope of AB =

Slope of the line perpendicular to AB = 1

Solution 6

Slope of the line passing through (0, 2) and (-3, -1) =

Slope of the line passing through (-1, 5) and (4, a) =

Since, the lines are parallel.

Solution 7

Slope of the line passing through (-4, -2) and (2, -3) =

Slope of the line passing through (a, 5) and (2, -1) =

Since, the lines are perpendicular.

Solution 8

The given points are A (4, -2), B (-4, 4) and C (10, 6).


 S l o p e space o f space A B equals fraction numerator 4 plus 2 over denominator minus 4 minus 4 end fraction equals fraction numerator 6 over denominator minus 8 end fraction equals fraction numerator minus 3 over denominator 4 end fraction

S l o p e space o f space B C equals fraction numerator 6 minus 4 over denominator 10 plus 4 end fraction equals 2 over 14 equals 1 over 7

S l o p e space o f space A C equals fraction numerator 6 plus 2 over denominator 10 minus 4 end fraction equals 8 over 6 equals 4 over 3

It can be seen that:

S l o p e space o f space A B equals fraction numerator minus 1 over denominator S l o p e space o f space A C end fraction


Hence, AB AC.


Thus, the given points are the vertices of a right-angled triangle.

Solution 9

The given points are A (4, 5), B (1, 2), C (4, 3) and D (7, 6).


S l o p e space o f space A B equals fraction numerator 2 minus 5 over denominator 1 minus 4 end fraction equals fraction numerator minus 3 over denominator minus 3 end fraction equals 1

S l o p e space o f space C D equals fraction numerator 6 minus 3 over denominator 7 minus 4 end fraction equals 3 over 3 equals 1


Since, slope of AB = slope of CD

 

Therefore AB || CD

 

S l o p e space o f space B C equals fraction numerator 3 minus 2 over denominator 4 minus 1 end fraction equals 1 third
S l o p e space o f space D A equals fraction numerator 5 minus 6 over denominator 4 minus 7 end fraction equals fraction numerator minus 1 over denominator minus 3 end fraction equals 1 third

 

 

Since, slope of BC = slope of DA


Therefore, BC || DA

 

Hence, ABCD is a parallelogram

Solution 10

Let the given points be A (-2, 4), B (4, 8), C (10, 7) and D (11, -5).


Let P, Q, R and S be the mid-points of AB, BC, CD and DA respectively.

 

Co-ordinates of P are



 

Co-ordinates of Q are



 

Co-ordinates of R are



 

Co-ordinates of S are



S l o p e space o f space P Q equals fraction numerator begin display style 15 over 2 end style minus 6 over denominator 7 minus 1 end fraction equals fraction numerator begin display style fraction numerator 15 minus 12 over denominator 2 end fraction end style over denominator 6 end fraction equals 3 over 12 equals 1 fourth
S l o p e space o f space R S equals fraction numerator begin display style fraction numerator minus 1 over denominator 2 end fraction end style minus 1 over denominator begin display style 9 over 2 end style minus begin display style 21 over 2 end style end fraction equals fraction numerator begin display style fraction numerator minus 1 minus 2 over denominator 2 end fraction end style over denominator begin display style fraction numerator 9 minus 21 over denominator 2 end fraction end style end fraction equals fraction numerator minus 3 over denominator minus 12 end fraction equals 1 fourth


 

 

Since, slope of PQ = Slope of RS, PQ || RS.


S l o p e space o f space Q R equals fraction numerator 1 minus begin display style 15 over 2 end style over denominator begin display style 21 over 2 end style minus 7 end fraction equals fraction numerator begin display style fraction numerator 2 minus 15 over denominator 2 end fraction end style over denominator begin display style fraction numerator 21 minus 14 over denominator 2 end fraction end style end fraction equals fraction numerator minus 13 over denominator 7 end fraction
S l o p e space o f space S P equals fraction numerator 6 plus begin display style 1 half end style over denominator 1 minus begin display style 9 over 2 end style end fraction equals fraction numerator begin display style fraction numerator 12 plus 1 over denominator 2 end fraction end style over denominator begin display style fraction numerator 2 minus 9 over denominator 2 end fraction end style end fraction equals fraction numerator 13 over denominator minus 7 end fraction equals fraction numerator minus 13 over denominator 7 end fraction

 

 

 

Since, slope of QR = Slope of SP, QR || SP.


Hence, PQRS is a parallelogram.

Solution 11

The points P, Q, R will be collinear if slope of PQ and QR is the same.

 

 

S l o p e space o f space P Q equals fraction numerator c plus a minus b minus c over denominator b minus a end fraction equals fraction numerator a minus b over denominator b minus a end fraction equals minus 1
S l o p e space o f space Q R equals fraction numerator a plus b minus c minus a over denominator c minus b end fraction equals fraction numerator b minus c over denominator c minus b end fraction equals minus 1


 

Hence, the points P, Q, and R are collinear.

Solution 12

Let A = (x, 2) and B = (8, -11)

Slope of AB =

Solution 13

We know that the slope of any line parallel to x-axis is 0.

Therefore, slope of AB = 0

 

Since, ABC is an equilateral triangle, 

Slope of AC = tan 60o =

 

Slope of BC = -tan 60o = -

Solution 14

We know that the slope of any line parallel to x-axis is 0.

Therefore, slope of AB = 0

 

As CD || BC, slope of CD = Slope of AB = 0

 

As BC AB, slope of BC =

As AD AB, slope of AD =

 

(i) The diagonal AC makes an angle of 45o with the positive direction of x axis.

(ii) The diagonal BC makes an angle of -45o with the positive direction of x axis.

Solution 15

Given, A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC.

 

(i) Slope of AB =

Slope of the altitude of AB =

 

(ii) Since, D is the mid-point of BC.

Co-ordinates of point D are

Slope of AD =

 

(iii) Slope of AC =

Slope of line parallel to AC = Slope of AC = 3

Solution 16

(i) Since, BC is perpendicular to AB,

Slope of AB =

 

(ii) Since, AD is parallel to BC,

Slope of AD = Slope of BC =

Solution 17

(i) A = (-3, -2) and B = (1, 2)

Slope of AB =

Inclination of line AB = = 45o

 

(ii) A = (0, ) and B = (3, 0)

Slope of AB =

Inclination of line AB = = 30o

 

(iii) A = (-1, 2) and B = (-2, )

Slope of AB =

Inclination of line AB = = 60o

Solution 18

Given, points A (-3, 2), B (2, -1) and C (a, 4) are collinear.

Slope of AB = Slope of BC

Solution 19

Given, points A (K, 3), B (2, -4) and C (-K + 1, -2) are collinear.

Slope of AB = Slope of BC

Solution 20

 

From the graph, clearly, AC has steeper slope.

 

Slope of AB =

Slope of AC =

The line with greater slope is steeper. Hence, AC has steeper slope.

Solution 21

Since, PQ || RS,

Slope of PQ = Slope of RS

 

(i) Slope of PQ =

Slope of RS =

 

(ii) Slope of PQ =

Slope of RS =

 

(iii) Slope of PQ =

Slope of RS =

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(C)

Solution 1

Given, y-intercept = c = 2 and slope = m = 3.

Substituting the values of c and m in the equation y = mx + c, we get,

y = 3x + 2, which is the required equation.

Solution 2

Given, y-intercept = c = -1 and inclination = 45o.

Slope = m = tan 45o = 1

Substituting the values of c and m in the equation y = mx + c, we get,

y = x - 1, which is the required equation.

Solution 3

Given, slope =

The equation passes through (-3, 4) = (x1, y1)

Substituting the values in y - y1 = m(x - x1), we get,

y - 4 = (x + 3)

3y - 12 = -4x - 12

4x + 3y = 0, which is the required equation.

Solution 4

Slope of the line = tan 60o =

The line passes through the point (5, 4) = (x1, y1)

Substituting the values in y - y1 = m(x - x1), we get,

y - 4 = (x - 5)

y - 4 = x - 5

y =x + 4 - 5, which is the required equation.

Solution 5

(i) Let (0, 1) = (x1, y1) and (1, 2) = (x2, y2)

The required equation of the line is given by:

y - y1 = m(x - x1)

y - 1 = 1(x - 0)

y - 1 = x

y = x + 1

(ii) Let (-1, -4) = (x1, y1) and (3, 0) = (x2, y2)

The required equation of the line is given by:

y - y1 = m(x - x1)

y + 4 = 1(x + 1)

y + 4 = x + 1

y = x - 3

Solution 6

Given, co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively.

(i) Gradient of PQ =

(ii) The equation of the line PQ is given by:

y - y1 = m(x - x1)

y - 6 = (x - 2)

5y - 30 = x - 2

5y = x + 28

(iii) Let the line PQ intersects the x-axis at point A (x, 0).

Putting y = 0 in the equation of the line PQ, we get,

0 = x + 28

x = -28

Thus, the co-ordinates of the point where PQ intersects the x-axis are A (-28, 0).

Solution 7

(i) Given, co-ordinates of two points A and B are (-3, 4) and (2, -1).

Slope =

The equation of the line AB is given by:

y - y1 = m(x - x1)

y + 1 = -1(x - 2)

y + 1 = -x + 2

x + y = 1

(ii) Let the line AB intersects the y-axis at point (0, y).

Putting x = 0 in the equation of the line, we get,

0 + y = 1

y = 1

Thus, the co-ordinates of the point where the line AB intersects the y-axis are (0, 1).

Solution 8

Slope of line AB = tan 45o = 1

The line AB passes through P (3, 4). So, the equation of the line AB is given by:

y - y1 = m(x - x1)

y - 4 = 1(x - 3)

y - 4 = x - 3

y = x + 1

Slope of line CD = tan 60o =

The line CD passes through P (3, 4). So, the equation of the line CD is given by:

y - y1 = m(x - x1)

y - 4 = (x - 3)

y - 4 = x - 3

y = x + 4 - 3

Solution 9

Solution 10

Since, ABCD is a parallelogram,

B = 180o - 60o = 120o

Slope of BC = tan 120o = tan (90o + 30o) = cot30o =

Equation of the line BC is given by:

y - y1 = m(x - x1)

y - 5 = (x - 7)

y - 5 = x - 7

y = x + 5 - 7

Since, CD || AB and AB || x-axis, slope of CD = Slope of AB = 0

Equation of the line CD is given by:

y - y1 = m(x - x1)

y - 5 = 0(x - 7)

y = 5

Solution 11

The given equations are:


x + 2y = 7 ....(1)


x - y = 4 ....(2)

 

Subtracting (2) from (1), we get,


3y = 3


y = 1

 

From (2), x = 4 + y = 4 + 1 = 5

 

The required line passes through (0, 0) and (5, 1).


S l o p e space o f space t h e space l i n e equals fraction numerator 1 minus 0 over denominator 5 minus 0 end fraction equals 1 fifth

 

Required equation of the line is given by:


y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
rightwards double arrow y minus 0 equals 1 fifth open parentheses x minus 0 close parentheses
rightwards double arrow 5 y equals x
rightwards double arrow x minus 5 y equals 0

Solution 12

Given, the co-ordinates of vertices A, B and C of a triangle ABC are (4, 7), (-2, 3) and (0, 1) respectively.

Let AD be the median through vertex A.

Co-ordinates of the point D are

Slope of AD =

The equation of the median AD is given by:

y - y1 = m(x - x1)

y - 2 = 1(x + 1)

y - 2 = x + 1

y = x + 3

The slope of the line which is parallel to line AC will be equal to the slope of AC.

Slope of AC =

The equation of the line which is parallel to AC and passes through B is given by:

y - 3 = (x + 2)

2y - 6 = 3x + 6

2y = 3x + 12

Solution 13

Slope of BC =

Slope of line perpendicular to BC =

The equation of the line through A and perpendicular to BC is given by:

y - y1 = m(x - x1)

y - 3 = 1(x - 0)

y - 3 = x

y = x + 3

Solution 14

Let A = (1, 4), B = (2, 3), and C = (-1, 2).

Slope of AB =

Slope of equation perpendicular to AB =

The equation of the perpendicular drawn through C onto AB is given by:

y - y1 = m(x - x1)

y - 2 = 1(x + 1)

y - 2 = x + 1

y = x + 3

Solution 15

(i) When x-intercept = 5, corresponding point on x-axis is (5, 0)

When y-intercept = 3, corresponding point on y-axis is (0, 3).

Let (x1, y1) = (5, 0) and (x2, y2) = (0, 3)

Slope =

 

The required equation is:

y - y1 = m(x - x1)

y - 0 = (x - 5)

5y = -3x + 15

3x + 5y = 15

 

(ii) When x-intercept = -4, corresponding point on x-axis is (-4, 0)

When y-intercept = 6, corresponding point on y-axis is (0, 6).

Let (x1, y1) = (-4, 0) and (x2, y2) = (0, 6)

Slope =

 

The required equation is:

y - y1 = m(x - x1)

y - 0 = (x + 4)

2y = 3x + 12

 

(iii) When x-intercept = -8, corresponding point on x-axis is (-8, 0)

When y-intercept = -4, corresponding point on y-axis is (0, -4).

Let (x1, y1) = (-8, 0) and (x2, y2) = (0, -4)

Slope =

 

The required equation is:

y - y1 = m(x - x1)

y - 0 = (x + 8)

2y = -x - 8

x + 2y + 8 = 0

Solution 16

Since, x-intercept is 6, so the corresponding point on x-axis is (6, 0).

Slope = m =

Required equation of the line is given by:

y - y1 = m(x - x1)

y - 0 = (x - 6)

6y = -5x + 30

5x + 6y = 30

Solution 17

Since, x-intercept is 5, so the corresponding point on x-axis is (5, 0).

The line also passes through (-3, 2).

Slope of the line =

Required equation of the line is given by:

y - y1 = m(x - x1)

y - 0 = (x - 5)

4y = -x + 5

x + 4y = 5

Solution 18

Since, y-intercept = 5, so the corresponding point on y-axis is (0, 5).

The line passes through (1, 3).

Slope of the line =

Required equation of the line is given by:

y - y1 = m(x - x1)

y - 5 = -2(x - 0)

y - 5 = -2x

2x + y = 5

Solution 19

Let AB and CD be two equally inclined lines.

For line AB:

Slope = m = tan 45o = 1

(x1, y1) = (-2, 0)

Equation of the line AB is:

y - y1 = m(x - x1)

y - 0 = 1(x + 2)

y = x + 2

For line CD:

Slope = m = tan (-45o) = -1

(x1, y1) = (-2, 0)

Equation of the line CD is:

y - y1 = m(x - x1)

y - 0 = -1(x + 2)

y = -x - 2

x + y + 2 = 0

Solution 20

(i)

 

 

(ii)

 

 

(iii)

Solution 21

Given, P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3: 1.

Co-ordinates of point P are

Slope = m = (Given)

Thus, the required equation of the line is

y - y1 = m(x - x1)

y + 2 = (x - 10)

5y + 10 = -2x + 20

2x + 5y = 10

Solution 22

(i) Co-ordinates of the centroid of triangle ABC are

(ii) Slope of AB =

Slope of the line parallel to AB = Slope of AB = -1

Thus, the required equation of the line is

y - y1 = m(x - x1)

Solution 23

Given, AP: CP = 2: 3

Co-ordinates of P are

Slope of BP =

Required equation of the line passing through points B and P is

y - y1 = m(x - x1)

y - 1 = 0(x - 3)

y = 1

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(D)

Solution 1

(i) y = 4

Comparing this equation with y = mx + c, we have:

Slope = m = 0

y-intercept = c = 4

(ii) ax - by = 0 by = ax y =

Comparing this equation with y = mx + c, we have:

Slope = m =

y-intercept = c = 0

(iii) 3x - 4y = 5

Comparing this equation with y = mx + c, we have:

Slope = m =

y-intercept = c =

Solution 2

Given equation of a line is x - y = 4

y = x - 4

Comparing this equation with y = mx + c. We have:

Slope = m = 1

y-intercept = c = -4

Let the inclination be .

Slope = 1 = tan = tan 45o

Solution 3

(i) 3x + 4y + 7 = 0

Slope of this line =

28x - 21y + 50 = 0

Slope of this line =

Since, product of slopes of the two lines = -1, the lines are perpendicular to each other.

(ii) x - 3y = 4

3y = x - 4

y =

Slope of this line =

3x - y = 7

y = 3x - 7

Slope of this line = 3

Product of slopes of the two lines = 1 -1

So, the lines are not perpendicular to each other.

(iii) 3x + 2y = 5

2y = -3x + 5

y =

Slope of this line =

x + 2y = 1

2y = -x + 1

y =

Slope of this line =

Product of slopes of the two lines = 3 -1

So, the lines are not perpendicular to each other.

(iv) Given, the slope of the line through (1, 4) and (x, 2) is 2.

Solution 4

(i) x + 2y + 3 = 0

2y = -x - 3

y =

Slope of this line =

Slope of the line which is parallel to the given line = Slope of the given line =

(ii)

Slope of this line =

Slope of the line which is parallel to the given line = Slope of the given line =

Solution 5

(i)

Slope of this line = 2

Slope of the line which is perpendicular to the given line =

(ii)

Slope of this line =

Slope of the line which is perpendicular to the given line =

Solution 6

(i) 2x - by + 3 = 0

by = 2x + 3

y =

Slope of this line =

ax + 3y = 2

3y = -ax + 2

y =

Slope of this line =

Since, the lines are parallel, so the slopes of the two lines are equal.


(ii) mx + 3y + 7 = 0

3y = -mx - 7

y =

Slope of this line =

5x - ny - 3 = 0

ny = 5x - 3

y =

Slope of this line =

Since, the lines are perpendicular; the product of their slopes is -1.

Solution 7

2x - y + 5 = 0

y = 2x + 5

Slope of this line = 2

px + 3y = 4

3y = -px + 4

y =

Slope of this line =

Since, the lines are perpendicular to each other, the product of the slopes is -1.

Solution 8

(i) 2x - 2y + 3 = 0

2y = 2x + 3

y = x +

Slope of the line AB = 1

(ii) Required angle =

Slope = tan = 1 = tan 45o

= 45o

Solution 9

4x + 3y = 9

3y = -4x + 9

y = + 3

Slope of this line =

px - 6y + 3 = 0

6y = px + 3

y =

Slope of this line =

Since, the lines are parallel, their slopes will be equal.

Solution 10

y = 3x + 7

Slope of this line = 3

2y + px = 3

2y = -px + 3

y =

Slope of this line =

Since, the lines are perpendicular to each other, the product of their slopes is -1.

Solution 11

 

 

 

Solution 12

(i) The slope of the line parallel to x-axis is 0.

(x1, y1) = (-5, 7)

Required equation of the line is

y - y1 = m(x - x1)

y - 7 = 0(x + 5)

y = 7

(ii) The slope of the line parallel to y-axis is not defined.

That is slope of the line is and hence the given line is parallel to y-axis.

(x1, y1) = (-5, 7)

Required equation of the line is

x - x1 =0

x + 5=0

Solution 13

(i) x - 3y = 4

3y = x - 4

Slope of this line =

Slope of a line parallel to this line =

Required equation of the line passing through (5, -3) is

y - y1 = m(x - x1)

y + 3 = (x - 5)

3y + 9 = x - 5

x - 3y - 14 = 0

(ii) 2y = -3x + 8

Or y =

Slope of given line =

Since the required line is parallel to given straight line.

Slope of required line (m) =

Now the equation of the required line is given by:

y - y1 = m(x - x1)

y - 1 =

2y - 2 = -3x

3x + 2y = 2

Solution 14

4x + 5y = 6

5y = -4x + 6

y =

Slope of this line =

The required line is perpendicular to the line 4x + 5y = 6.

The required equation of the line is given by

y - y1 = m(x - x1)

y - 1 = (x + 2)

4y - 4 = 5x + 10

5x - 4y + 14 = 0

Solution 15

Let A = (6, -3) and B = (0, 3).

We know the perpendicular bisector of a line is perpendicular to the line and it bisects the line, that it, it passes through the mid-point of the line.

Co-ordinates of the mid-point of AB are

Thus, the required line passes through (3, 0).

Slope of AB =

Slope of the required line =

Thus, the equation of the required line is given by:

y - y1 = m(x - x1)

y - 0 = 1(x - 3)

y = x - 3

Solution 16

(i) The co-ordinates of points A, B and C are (2, 3), (-1, 2) and (3, 0) respectively.

(ii) Slope of BC =

Slope of a line parallel to BC = Slope of BC =

Required equation of a line passing through A and parallel to BC is given by

y - y1 = m(x - x1)

y - 3 = (x - 2)

2y - 6 = -x + 2

X + 2y = 8

Solution 17

We know that in a rhombus, diagonals bisect each other at right angle.

Let O be the point of intersection of the diagonals AC and BD.

Co-ordinates of O are

Slope of BD =

For line BD:

Slope = m = , (x1, y1) = (-5, 6)

Equation of the line BD is

y - y1 = m(x - x1)

y - 6 = (x + 5)

3y - 18 = -x - 5

x + 3y = 13

For line AC:

Slope = m = , (x1, y1) = (-2, 5)

Equation of the line AC is

y - y1 = m(x - x1)

y - 5 = 3(x + 2)

y - 5 = 3x + 6

y = 3x + 11

Solution 18

We know that in a square, diagonals bisect each other at right angle.

Let O be the point of intersection of the diagonals AC and BD.

Co-ordinates of O are

Slope of AC =

For line AC:

Slope = m = , (x1, y1) = (7, -2)

Equation of the line AC is

y - y1 = m(x - x1)

y + 2 = (x - 7)

2y + 4 = x - 7

2y = x - 11

For line BD:

Slope = m = , (x1, y1) = (3, -4)

Equation of the line BD is

y - y1 = m(x - x1)

y + 4 = -2(x - 3)

y + 4 = -2x + 6

2x + y = 2

Solution 19

(i) We know the median through A will pass through the mid-point of BC. Let AD be the median through A.

Co-ordinates of the mid-point of BC, i.e., D are

Slope of AD =

Equation of the median AD is

y - 3 = -8(x - 0)

8x + y = 3

(ii) Let BE be the altitude of the triangle through B.

Slope of AC =

Slope of BE =

Equation of altitude BE is

y - 2 = (x - 2)

3y - 6 = x - 2

3y = x + 4

(iii) Slope of AB =

Slope of the line parallel to AB = Slope of AB = 7

So, the equation of the line passing through C and parallel to AB is

y - 4 = 7(x + 2)

y - 4 = 7x + 14

y = 7x + 18

Solution 20

(i) 2y = 3x + 5

Slope of this line =

Slope of the line AB =

(x1, y1) = (3, 2)

The required equation of the line AB is

y - y1 = m(x - x1)

y - 2 = (x - 3)

3y - 6 = -2x + 6

2x + 3y = 12

(ii) For the point A (the point on x-axis), the value of y = 0.

2x + 3y = 12 2x = 12 x = 6

Co-ordinates of point A are (6, 0).

For the point B (the point on y-axis), the value of x = 0.

2x + 3y = 12 3y = 12 y = 4

Co-ordinates of point B are (0, 4).

Area of OAB = OA OB = 6 4 = 12 sq units

Solution 21

For the point A (the point on x-axis), the value of y = 0.

4x - 3y + 12 = 0 4x = -12 x = -3

Co-ordinates of point A are (-3, 0).

Here, (x1, y1) = (-3, 0)

The given line is 4x - 3y + 12 = 0

3y = 4x + 12

y = + 4

Slope of this line =

Slope of a line perpendicular to the given line =

Required equation of the line passing through A is

y - y1 = m(x - x1)

y - 0 = (x + 3)

4y = -3x - 9

3x + 4y + 9 = 0

Solution 22

(i) The given equation is

2x - 3y + 18 = 0

3y = 2x + 18

y = x + 6

Slope of this line =

Slope of a line perpendicular to this line =

(x1, y1) = (-5, 7)

The required equation of the line AP is given by

y - y1 = m(x - x1)

y - 7 = (x + 5)

2y - 14 = -3x - 15

3x + 2y + 1 = 0

(ii) P is the foot of perpendicular from point A.

So P is the point of intersection of the lines 2x - 3y + 18 = 0 and 3x + 2y + 1 = 0.

2x - 3y + 18 = 0 4x - 6y + 36 = 0

3x + 2y + 1 = 0 9x + 6y + 3 = 0

Adding the two equations, we get,

13x + 39 = 0

x = -3

3y = 2x + 18 = -6 + 18 = 12

y = 4

Thus, the co-ordinates of the point P are (-3, 4).

Solution 23

For the line AB:

S l o p e space o f space A B equals m equals fraction numerator 2 minus 0 over denominator 2 minus 4 end fraction equals fraction numerator 2 over denominator minus 2 end fraction equals minus 1

(x1, y1) = (4, 0)

 

Equation of the line AB is

y - y1 = m(x - x1)

y - 0 = -1(x - 4)

y = -x + 4

x + y = 4 ....(1)

 

For the line BC:

S l o p e space o f space B C equals m equals fraction numerator 6 minus 2 over denominator 0 minus 2 end fraction equals fraction numerator 4 over denominator minus 2 end fraction equals minus 2

(x1, y1) = (2, 2)

 

Equation of the line BC is

y - y1 = m(x - x1)

y - 2 = -2(x - 2)

y - 2 = -2x + 4

2x + y = 6 ....(2)

 

Given that AB cuts the y-axis at P. So, the abscissa of point P is 0.

Putting x = 0 in (1), we get,

y = 4

 

Thus, the co-ordinates of point P are (0, 4).

 

Given that BC cuts the x-axis at Q. So, the ordinate of point Q is 0.

Putting y = 0 in (2), we get,

2x = 6 x = 3

 

Thus, the co-ordinates of point Q are (3, 0).

Solution 24

Putting x = 0 and y = 0 in the equation y = 2x, we have:

LHS = 0 and RHS = 0

Thus, the line y = 2x passes through the origin.

Hence, A = L3

Putting x = 0 in y - 2x + 2 = 0, we get, y = -2

Putting y = 0 in y - 2x + 2 = 0, we get, x = 1

So, x-intercept = 1 and y-intercept = -2

So, x-intercept is positive and y-intercept is negative.

Hence, B = L4

Putting x = 0 in 3x + 2y = 6, we get, y = 3

Putting y = 0 in 3x + 2y = 6, we get, x = 2

So, both x-intercept and y-intercept are positive.

Hence, C = L2

The slope of the line y = 2 is 0.

So, the line y = 2 is parallel to x-axis.

Hence, D = L1

Solution 25

TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 14 - Equation of a Line  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well. 

Text Book Solutions

ICSE X - Mathematics

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