Selina Concise Solution Class 10 Chapter 14 Equation of a ...

Selina Concise Solution Class 10

Chapter 14

Equation of a Line

https://www.swiflearn.com/

EXERCISE 14E

1. Point P divides the line segment joining the points A (8, 0) and B (16, -8) in

the ratio 3: 5. Find its co-ordinates of point P.

Also, find the equation of the line through P and parallel to 3x + 5y = 7.

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

( )( )

( )1 1

3 8 5 03 16 5 8, 11, 3

3 5 3 5

,x y

− + + = −

+ +

=

Given, 3 5 7

3 7

5 5

x y

y x

+ =

−= +

3Slope of the line =

5

−

Thus the equation of the required line:

( )3

3 115

5 15 3 33

3 5 18

y x

y x

x y

−+ = −

+ = − +

+ =

2. The line segment joining the points A(3, -4) and B (-2, 1) is divided in the

ratio 1: 3 at point P in it. Find the co-ordinates of P. Also, find the equation of

the line through P and perpendicular to the line 5x – 3y + 4 = 0.

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

( ) ( )

( )1 1

1 2 3 3 1 1 3 4 7 11, ,

1 3 1 3 4 4

,x y

− + + − − =

+ +

=

Given, 5 3 4

5 4

3 3

x y

y x

− =

= −


3

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



Chapter 14

Equation of a Line


1Slope of the required line =

5

3

3

5

−

−=


11 3 7

4 5 4

4 11 3 4 7

4 5 4

20 55 12 21

12 20 34 0

6 10 17 0

y x

y x

y x

x y

x y

− + = −

+ − − =

+ = − +

+ + =

+ + =

3. A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of

point P. Find the equation of a line through P and perpendicular to x – 3y + 4

= 0.

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

y = -5

Thus, the co-ordinates of the point P are (0, -5). 3 4 0

1 4

3 3

x y

y x

− + =

= +


3

Since, the required line is perpendicular to the given equation x - 3y + 4 = 0.,


1

3

3

−

= −

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


( )5 3 0

3 5 0

y x

x y

+ = − −

+ + =

4. Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0

are perpendicular to each other.



Chapter 14

Equation of a Line


Solution: Given, 5 4

4

5 5

y kx

ky x

= +

= +

1Slope of this line, m5

k=

5 2 5 0

2 5 5

5 5

2 2

x y

y x

y x

− + =

= +

= +

2

5Slope of this line, m

2=

Since, the lines are perpendicular, m1m2 = -1

51

5 2

2

k

k

= −

= −

5. A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects

the co-ordinate axes at points A and B. M is the mid-point of the segment AB.

Find:

(i) The equation of the line.

(ii) The co-ordinates of A and B.

(iii) The co-ordinates of M.

Solution: (i)



Chapter 14

Equation of a Line


2 4Slope of PQ

5 1

6

6

1

− −=

+

−=

= −

Equation of the line PQ is given by:

( )4 1 1

4 1

3

y x

y x

x y

− = − +

− = − −

+ =

(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:

3 0 0 3 3 3, ,

2 2 2 2

+ + =

6. (1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of

rhombus ABCD. Find the equations of the diagonals AC and BD.

Solution: 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:

( )1 3 5 1

, 1,22 2

− − = −



Chapter 14

Equation of a Line


1 5Slope of AC =

3 1

6

4

3

2

− −

− −

−=−

=

For line AC:

( ) ( )1 1

3Slope, m = , , 1,5

2x y =

Equation of the line AC is:

( )3

5 12

2 10 3 3

3 2 7 0

y x

y x

x y

− = −

− = −

− + =

For line BD:

( ) ( )1 1

1Slope, m =

Slope of AC

2

3

, 1,2x y

−

−=

= −

Equation of the line BD is:

( )2

2 13

3 6 2 2

2 3 4

y x

y x

x y

−− = +

− = − −

+ =

7. Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.

(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.

(ii) Without using the co-ordinates of vertex D, find the equation of side AD of

the square and also the equation of diagonal BD.

Solution: Using distance formula:



Chapter 14

Equation of a Line


( ) ( )

( ) ( )

2 2

2 2

AB = 6 3 2 2

9 16

5

BC = 2 6 5 2

16 9

5

− + − −

= +

=

− + − +

= +

=

Thus, AC= BC.

Also,

2 2Slope of AB =

6 3

4

3

5 2Slope of BC =

2 6

3

4

3

4

Slope of AB Slope of BC = 1

− −

−

−=

− +

−

−=−

=

−

Thus, AB ⊥ BC

Hence, A, B, C can be the vertices of a square.

(i)

2 2Slope of AB =

6 3

Slope of CD

− −

−

=

Equation of the line CD :

( )4

5 23

3 15 4 8

4 3 7 ...(1)

y x

y x

x y

−+ = −

+ = − +

+ = −

5 2Slope of BC =

2 6

3

4

3

4

Slope of AD

− +

−

−=−

=

=



Chapter 14

Equation of a Line


Equation of the line AD :

( )3

2 34

4 8 3 9

3 4 1 ...(2)

y x

y x

x y

− = −

− = −

− =

Now, D is the point of intersection of CD and AD.

Multiply the equation (1) by 4 and multiply the equation (1) by 2. 16 12 28...(3)

9 12 3 ...(4)

x y

x y

+ = −

− =

Adding the equations (3) and (4),

25 25

1

x

x

= −

= −

Thus, 4 3 1

3 1

4 4

1

y x

y

y

= −

= − −

= −

= −

Thus, the co-ordinates of point D are (-1, -1).

(ii)

The equation of line AD is found in part (i), it is 3x – 4y = 1 or 4y = 3x – 1.

1 2Slope of BD =

1 6

1

7

1

7

− +

− −

=−

−=

The equation of diagonal BD is:

( )1

1 17

7 7 1

7 8 0

y x

y x

x y

−+ = +

+ = − −

+ + =

8. A line through origin meets the line x = 3y + 2 at right angles at point X.

Find the co-ordinates of X.

Solution: The given line is:



Chapter 14

Equation of a Line


3 2 ...(1)

3 2

1 2

3 3

x y

y x

y x

= +

= −

= −


3

The required line intersects the given line at right angle.


1

3

3

−

= −

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

The equation of the required line is:

( )0 3 0

3 0 ...(2)

y x

x y

− = − −

+ =

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

Using (1) in (2), 9 6 0

6

10

3

5

y y

y

+ + =

−=

−=

Therefore, 3 2

92

5

1

5

x y= +

−= +

=

Thus, the co-ordinates of the point X are 1 3

,5 5

−

.

9. A straight line passes through the point (3, 2) and the portion of this line,

intercepted between the positive axes, is bisected at this point. Find the

equation of the line.

Solution:



Chapter 14

Equation of a Line


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:

( )

( )

0 0, 3,2

2 2

, 3,22 2

6, 4

x y

x y

x y

+ + =

=

= =

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

4 0Slope of line AB =

0 6

4

6

2

3

−

−

=−

−=

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

The required equation of the line AB is given by:

( )2

0 63

3 2 12

2 3 12

y x

y x

x y

−− = −

= − +

+ =

10. Find the equation of the line passing through the point of intersection of

7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x – 2y = 1.

Solution: 7x + 6y = 71

28x + 24y = 284 ... (1)

5x - 8y = -23

15x - 24y = -69 ... (2)

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

43x = 215

x = 5

From (2),



Chapter 14

Equation of a Line


8y = 5x + 23

= 25 + 23

= 48

y = 6

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

4 2 1

2 4 1

12

2

x y

y x

y x

− =

= −

= −

Slope of this line = 2

1Slope of the required line

2

−=

The required equation of the line is:

( )1

6 52

2 12 5

2 17

y x

y x

x y

−− = −

− = − +

+ =

11. Find the equation of the line which is perpendicular to the line 1x y

a b− = at

the point where this line meets y-axis.

Solution: The given line is:

1

1

x y

a b

y x

b a

by x b

a

− =

= −

= −

Slope of this line b

a=

1Slope of the required line

b

a

a

b

−=

−=

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



Chapter 14

Equation of a Line


Putting x = 0 in the equation 1x y

a b− = ,

0 1y

b

y b

− =

= −

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

The equation of the required line is:

( )

2

2

0

0

ay b x

b

by b ax

ax by b

−+ = −

+ = −

+ + =

12. O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:

(i) the equation of median of triangle OAB through vertex O.

(ii) the equation of altitude of triangle OAB through vertex B.

Solution: (i) Let the median through O meets AB at D.

So, D is the mid-point of AB.

Co-ordinates of point D are:

( )3 5 5 3

, 1,12 2

− − = −

1 0Slope of OD =

1 0

1

−

− −

= −

(x1, y1) = (0, 0)

The equation of the median OD is:

( )0 1 0

0

y x

x y

− = − −

+ =

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

5 0Slope of OA =

3 0

5

3

−

−

=

1Slope of the required altitude =

5

3

3

5

−

−=

The equation of the required altitude through B is:



Chapter 14

Equation of a Line


( )3

3 55

5 15 3 15

3 5 30 0

y x

y x

x y

−+ = +

+ = − −

+ + =

13. Determine whether the line through points (-2, 3) and (4, 1) is

perpendicular to the line 3x = y + 1.

Does the line 3x = y + 1 bisect the line segment joining the two given points?

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

1

1 3Slope of AB, m =

4 2

2

6

1

3

−

+

−=

−=

Equation of line AB:

( )1

3 23

2 9 2

2 7 .....(1)

y x

y x

x y

−− = +

− = − −

+ −

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

m1 × m2 = -1

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),

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:

( )2 4 3 1

, 1,22 2

P

− + + =

=

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

14. Given a straight line cos30 sin30 2x y+ = . Determine the equation of the

other line which is parallel to it and passes through (4, 3).



Chapter 14

Equation of a Line


Solution: Given,

cos30 sin 30 2

3 12

2 2

3 4

3 4

x y

x y

x y

y x

+ =

+ =

+ =

= − +

Slope of this line = 3

Slope of a line which is parallel to this given line = 3

−

−

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

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

( )3 3 4

3 4 3 3

y x

x y

− = − −

+ = +

15. Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:

(i) perpendicular to the line 2x – y + 7 = 0

(ii) parallel to it.

Solution: Given,

( ) ( )

( ) ( )

2 3 5 0 ...(1)

3 2 5

2 5

3 3

k x k y

k y k x

ky x

k k

− + + − =

+ = − − +

− = +

+ +

1

2Slope of this line, m =

3

k

k

−

+

(i)

2x - y + 7 = 0

y = 2x + 7

Slope of this line, m2 = 2

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

( )

1 2m m 1

22 1

3

4 2 3

7

k

k

k k

k

= −

− = −

+

− = − −

=



Chapter 14

Equation of a Line


(ii)

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

1 2m m

22

3

2 2 6

3 4

4

3

k

k

k k

k

k

=

− =

+

− = +

= −

−=

16. The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write

down the equation of BC. Find:

(i) the equation of line through A and perpendicular to BC.

(ii) the co-ordinates of the point, where the perpendicular through A, as

obtained in (i), meets BC.

Solution: 7 2

Slope of BC = 11 1

9

12

3

4

+

+

=

=

Equation of the line BC is given by:

( )3

2 14

4 8 3 3

3 4 5 ......(1)

y x

y x

x y

+ = +

+ = +

− =

(i)

1Slope of line perpendicular to BC =

3

4

4

3

−

−=

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

( )4

5 03

3 15 4

4 3 15 ...(2)

y x

y x

x y

−− = −

− = −

+ =



Chapter 14

Equation of a Line


(ii)

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

Multiply equation (1) by 3 and multiply the equation (2) by 4.

9x - 12y = 15 … (3)

16x + 12y = 60 … (4)

Adding the equations (3) and (4),

25x = 75

x = 3

So,

4y = 3x - 5

= 9 - 5

4y = 4

y = 1

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

17. From the given figure, find:

(i) The co-ordinates of A, B and C.

(ii) The equation of the line through A and parallel to BC.

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

(ii)



Chapter 14

Equation of a Line


0 2Slope of BC =

3 1

2

4

1

2

−

+

−=

−=

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

1

2

−=

And (x1, y1) = (2, 3).

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

( )1

3 22

2 6 2

2 8

y x

y x

x y

−− = −

− = − +

+ =

18. P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write

down the equation of the median of the triangle through R.

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

The co-ordinates of point X are:

( )3 7 4 2

, 5,12 2

+ − =

1 1Slope of RX, m =

5 2

2

7

+

+

=

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

The required equation of the median RX is given by:

( )2

1 27

7 7 2 4

7 2 3

y x

y x

y x

+ = +

+ = +

= −

19. A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the

mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to

show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.

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



Chapter 14

Equation of a Line


( )8 4 6 2

, 2, 22 2

− − + = −

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

( )8 0 6 10

, 4, 82 2

+ − − = −

8 2Slope of PQ =

4 2

6

2

3

10 2Slope of BC =

0 4

12

4

3

− +

−

−=

= −

− −

+

−=

= −

Since, slope of PQ = Slope of BC,

PQ || BC

Also, we have:

2 2Slope of PB =

2 4

4

6

2

3

8 10Slope of QC =

4 0

1

2

− −

+

−=

−=

− +

−

=

Thus, PB is not parallel to QC.

Hence, PBCQ is a trapezium.

20. A line AB meets the x-axis at point A and y-axis at point B. The point P (-

4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:

(i) the co-ordinates of A and B.

(ii) the equation of line through P and perpendicular to AB.

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

Using section formula, we have:



Chapter 14

Equation of a Line


( )

( )

1 0 2 1 2 04, 2 ,

1 2 1 2

24, 2 ,

3 3

24 2

3 3

6 6

x y

x y

x y

x y

+ + − − =

+ +

− − =

− = − =

= − = −

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

(ii)

6 0Slope of AB =

0 6

6

6

1

− −

+

−=

= −

1Slope of the required line perpendicular to AB =

1

1

−

−

=

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

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

( )2 1 4

2 4

2

y x

y x

y x

+ = +

+ = +

= +

21. A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units

from the positive side of y-axis. Find the equation of the line.

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

Also, y-intercept = 3

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

3 0Slope of AB =

0 2

3

2

−

+

=

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

Required equation of the line AB is given by:

( )3

0 22

2 3 6

y x

y x

− = +

= +



Chapter 14

Equation of a Line


22. Find the equation of a line passing through the point (2, 3) and having the

x-intercept of 4 units.

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

Also, x-intercept = 4

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

0 3Slope of AB =

4 2

3

2

−

−

−=

And (x1, y1) = (4, 0).

Required equation of the line AB is given by:

( )3

0 42

2 3 12

3 2 12

y x

y x

x y

−− = −

= − +

+ =

23. The given figure (not drawn to scale) shows two straight lines AB and CD.

If equation of the line AB is: y = x + 1 and equation of the line CD is: y = 3 x

– 1. Write down the inclination of lines AB and CD; also, find the angle

between AB and CD.

Solution: Equation of the line AB is y = x + 1

Slope of AB = 1

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

So, RPQ = 45

Equation of line CD:



Chapter 14

Equation of a Line


3 1y x= −

Slope of CD = 3

Inclination of line CD = 60 (Since, tan 60 = 3)

So,

DQX = 60

DQX = 180 60

120

−

=

Using angle sum property in ΔPQR,

= 180 45 120

15

− −

=

24. Write down the equation of the line whose gradient is 3

2 and which passes

through P, where P divides the line segment joining A (2, 6) and B (3, -4) in

the ratio 2 : 3.

Solution: 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:

( ) ( )

( )

( )1 1

2 3 3 2 2 4 3 6 6 6 8 18, ,

2 3 2 3 5 5

0,2

,x y

+ − − + − − + =

+ +

=

=


2

The required equation of the line is given by:

( )3

2 02

2 4 3

2 3 4

y x

y x

y x

− = −

− =

= +

25. The ordinate of a point lying on the line joining the points (6, 4) and (7, -5)

is -23. Find the co-ordinates of that point.

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

5 4Slope of the line AB =

7 6

9

− −

−

= −

And (x1, y1) = (6, 4).



Chapter 14

Equation of a Line


The equation of the line AB is given by:

( )4 9 6

4 9 54

9 58 ......(1)

y x

y x

x y

− = − −

− = − +

+ =

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

Putting y = -23 in (1),

9 23 58

9 81

9

x

x

x

− =

=

=

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

26. Points A and B have coordinates (7, -3) and (1, 9) respectively. Find:

(i) the slope of AB.

(ii) the equation of the perpendicular bisector of the line segment AB.

(iii) the value of ‘p’ if (-2, p) lies on it.

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

(i)


1 7

2

+

−

= −

(ii)

1Slope of perpendicular bisector =

2

1

2

−

−

=

Mid-point of AB:

( )7 1 3 9

, 4,32 2

+ − + =

Equation of perpendicular bisector is:

( )1

3 42

2 6 4

2 2 0

y x

y x

x y

− = −

− = −

− + =

(iii)

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



Chapter 14

Equation of a Line


-2 - 2p + 2 = 0

2p = 0

p = 0

27. A and B are two points on the x-axis and y-axis respectively. P (2, -3) is the

mid-point of AB. Find the:

(i) Coordinates of A and B.

(ii) Slope of line AB.

(iii) Equation of line AB.

Solution: (i)

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

Mid-point of A and B is given by:

( )

0 0, ,

2 2 2 2

2, 3 ,2 2

2 and 32 2

4 and 6

x y x y

x y

x y

x y

+ + =

− =

= = −

= = −

So, A = (4, 0) and B = (0, -6)

(ii)


0 4

3

2

− −

−

=

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



Chapter 14

Equation of a Line


( )3

0 42

2 3 12

y x

y x

− = −

= −

28. The equation of a line 3x + 4y – 7 = 0. Find:

(i) the slope of the line.

(ii) the equation of a line perpendicular to the given line and passing through

the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.

Solution: Given, 3 4 7 0 ...(1)

4 3 7

3 7

4 4

x y

y x

y x

+ − =

= − +

−= +

(i)


4

−

(ii)

1Slope of the line perpendicular to the given line =

3

4

4

3

−

−

=

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 4

3 passes through point (2, 4).

Thus, the required equation of the line:

( )4

4 23

3 12 4 8

4 3 4 0

y x

y x

x y

− = −

− = −

− + =

29. ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find:

(i) Co-ordinates of A

(ii) Equation of diagonal BD

Solution:



Chapter 14

Equation of a Line


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:

2 5 4 8 7 4, ,

2 2 2 2

7,2

2

+ − + =

=

(i)

Since O is the midpoint of AC also,

4 7

2 2

2 8 14

2 6

3

x

x

x

x

+=

+ =

=

=

And

72

2

7 4

3

y

y

y

+=

+ =

= −

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

(ii)

Equation of BD:



Chapter 14

Equation of a Line


( )

( )

( )

2 11 1

2 1

8 44 2

5 2

124 2

3

4 4 8

4 12

y yy y x x

x x

y x

y x

y x

x y

−− = −

−

+ + = −

−

+ = −

+ = −

− =

30. Given equation of the line L1 is y = 4.

(i) Write the slope of the line L2 if L2 is the bisector of angle O.

(ii) Write the coordinates of point P.

(iii) Find the equation of L2.

Solution: (i)

Equation of line L1 is y = 4.

As L2 is the bisector of angle O. POX = 45

Slope = tan 45

1

=

Let co-ordinates of P be (x, y).



Chapter 14

Equation of a Line


(ii)

2 12

2 1

Slope of L

4 01

0

41

4

y y

x x

x

x

x

−=

−

−=

−

=

=

Co-ordinates of P are (4, 4).

(iii)

Equation of L2 is:

( )4 1 4

4 4

y x

y x

x y

− = −

− = −

=

31. (i) Equation of AB.

(ii) Equation of CD.



Chapter 14

Equation of a Line


Solution: (i)

( )3 4

Slope of AB3 5

1

8

−=

− −

−

Equation of AB is given by:

( )( )1

4 58

8 32 5

8 27

y x

y x

x y

−− = − −

− = − −

+ =

(ii)

AB and CD are perpendicular to each other. Thus, product of their slopes = -1. Slope of AB Slope of CD = 1

1Slope of CD = 1

8

Slope of CD 8

−

− −

=

Now, from the graph co-ordinates of D = (-3, 0).

Equation of line CD is given by:

( )0 8 3

8 24

y x

y x

− = +

= +

32. Find the equation of the line that has x-intercept = -3 and is perpendicular

to 3x + 5y = 1.

Solution: Slope of 3x + 5y = 1 is given by,



Chapter 14

Equation of a Line


3m =

5

−

1Slope of the line perpendicular to the given line =

3

5

5

3

−

−

=

And x- intercept = -3

Therefore,

( )

m c

50 3 c

3

c = 5

y x= +

= − +

Thus equation of the required line,

55

3

3 5 15

5 3 15 0

y x

y x

x y

= +

= +

− + =

33. A straight line passes through the points P(-1, 4) and Q(5, -2). It intersects

x-axis at point A and y-axis at point B. M is the mid- t point of the line

segment AB. Find:

(i) the equation of the line.

(ii) the co-ordinates of points A and B.

(iii) the co-ordinates of point M

Solution: (i)

The equation of the line passing through the points P(-1, 4) and Q(5, -2):

( )

( )( )( )

( )

2 11 1

2 1

2 44 1

5 1

64 1

6

4 1

3

y yy y x x

x x

y x

y x

y x

x y

−− = −

−

− −− = − − − −

−− = +

− = − −

+ =

(ii)

The line x + y = 3 cuts x-axis at point A.

Hence, its y-coordinate is 0. And, x-coordinate is given by:



Chapter 14

Equation of a Line


x + 0 = 3

x = 3

So, the coordinates of A are (3, 0).

The line x + y - 3 cuts y-axis at point B. Hence, its x-coordinate is 0.

And, y-coordinate is given by:

0 + y = 3 y = 3

So, the coordinates of B are (0, 3).

(iii)

Since M is the mid-point of line segment AB, so, coordinates of M:

3 0 0 3 3 3, ,

2 2 2 2

+ + =

34. In the given figure, line AB meets y-axis at point A. Line through C (2, 10)

and D intersects line AB at right angle at point R. Find:

(i) Equation of line AB.

(ii) Equation of line CD.

(iii) Co-ordinates of points E and D.

Solution: (i)

8 6Slope of AB, m

6 0

2

6

1

3

−=− −

=−

−=

The y-intercept of the line AB is 6.

Thus, the equation of the given line is given by the slope-intercept form, y = mx + c



Chapter 14

Equation of a Line


16

3

3 18

3 18

y x

y x

x y

= − +

= − +

+ =

(ii)

Since AB and CD intersect at the right angles,

AB CD

CD

CD

m m 1

1m 1

3

m 3

= −

− = −

=

Thus, the equation of the given line is given by the slope-point form,

( )

( )

1 1m

10 3 2

3 4 0

y y x x

y x

x y

− = −

− = −

− + =

(iii)

Since point E satisfies the equation of AB, and the y-coordinate of E is 0, we can find the x-

coordinate of E.

( )3 0 4 0

3 4

4

3

x

x

x

+ + =

= −

−=

So, the co-ordinates of D are 4

,03

−

.

35. A line through point P(4, 3) meets x-axis at point A and the y-axis at point

B. If BP is double of PA, find the equation of AB.

Solution: Since a line through point P meets x-axis at point A and y-axis at point B, coordinates of A

are(x, 0) and coordinates of B are (4, 3). BP = 2PA

BP 2

PA 1=

P divides AB in the ratio 2 : 1.

So, the coordinates of P are:

2 1 0 2 0 1 2, ,

2 1 2 1 3 3

x y x y + + =

+ +

But, coordinates of P are (4, 3).



Chapter 14

Equation of a Line


24 and 3

3 3

2 12 and 9

6 and 9

x y

x y

x y

= =

= =

= =

Co-ordinates of A are (6, 0) and co-ordinates of B are (0, 9).

9 0Slope of line AB =

0 6

9

6

3

2

−

−

=−

−=

Thus, the equation of line AB is given by:

( )3

0 62

2 3 18

3 2 18

y x

y x

x y

−− = −

= − +

+ =

36. Find the equation of line through the intersection of lines 2x – y = 1 and 3x

+ 2y = -9 and making an angle of 30° with positive direction of x-axis.

Solution: Since the line passing through the x-axis makes an angle of 30° with the positive direction of the

x-axis, the slope of the line is given by:

1tan 30

3 =

The intersection of the lines 2x - y = 1 and 3x + 2y = -9 is given by solving the equations

simultaneously.

So, multiplying equation 2x - y = 1 by 2,

4x - 2y = 2

Now add this resultant to the second equation 3x + 2y = -9.

Thus,

7x = -7

x = -1

Substituting the value of x in 2x - y = 1, we get y = -3.

Thus, the intersection of the lines is (-1, -3).

To find the equation of the required line, we use the slope-point form.



Chapter 14

Equation of a Line


( ) ( )( )

( )

13 1

3

13 1

3

13

3 3

y x

y x

xy

− = − −

+ = +

= + −

37. Find the equation of the line through the Points A(-1, 3) and B(0, 2).

Hence, show that the points A, B and C(1, 1) are collinear.

Solution:

( )2 3

Slope of line AB = 0 1

1

1

1

−

− −

−=

= −

Using the slope-point form, the equation of line AB is given by:

( )( )3 1 1

3 1

2

y x

y x

x y

− = − − −

− = − −

+ =

Now,

1 2Slope of line BC =

1 0

1

1

1

−

−

−=

= −

Since, slope of line AB = Slope of line BC, points A, B and C are collinear.

38. Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5,

10) and C(3, 2), find :

(i) the co-ordinates of the fourth vertex D.

(ii) length of diagonal BD.

(iii) equation of side AB of the parallelogram ABCD.

Solution: (i)

Let (x, y) be the co-ordinates of D.

Since the diagonals of a parallelogram bisect each other.

Therefore, mid-point of diagonal AC = Mid-point of diagonal BD



Chapter 14

Equation of a Line


( )

3 3 6 2 5 10, ,

2 2 2 2

5 103,4 ,

2 2

5 103 and 4

2 2

5 6 and 10 8

1 and 2

x y

x y

x y

x y

x y

+ + + + =

+ + =

+ += =

+ = + =

= = −

Therefore, the co-ordinates of D are (1, -2).

(ii)

( ) ( )

( ) ( )

2 2

2 2

Length of diagonal BD = 1 5 2 10

4 12

16 144

4 10

− + − −

= − + −

= +

=

(iii)

10 6Slope of side AB =

5 3

4

2

2

−

−

=

=

Thus, the equation of side AB is given by:

( )6 2 3

6 2 6

2 0

2

y x

y x

x y

y x

− = −

− = −

− =

=

39. In the figure given, ABC is a triangle and BC is parallel to the y-axis. AB

and AC intersect the y-axis at P and Q respectively.



Chapter 14

Equation of a Line


(i) Write the co-ordinates of A.

(ii) Find the length of AB and AC.

(iii) Find the ratio in which Q divides AC.

(iv) Find the equation of the line AC.

Solution: (i)

The line intersects the x-axis where y = 0.

Hence, the co-ordinates of A are (4, 0).

(ii)

( )( ) ( )

( )( ) ( )

2 2

2 2

Length of AB = 4 2 0 3

36 9

45

3 5 units

Length of AC = 4 2 0 4

36 16

52

2 13 units

− − + −

= +

=

=

− − + +

= +

=

=

(iii)

Let k : 1 be the required ratio which divides the line segment joining the co-ordinates A(4, 0)

and C(-2, -4).

Let the co-ordinates of Q be x and y.

( ) ( ) ( )2 1 4 4 0and

1 1

k kx y

k k

− + − += =

+ +

Q lies on the y-axis where x = 0.



Chapter 14

Equation of a Line


2 40

1

2 4 0

4

2

2

k

k

k

k

k

− +=

+

− + =

=

=

Thus, the required ratio is 2 : 1.

(iv)

4 0Slope of line AC =

2 4

4

6

2

3

− −

− −

−=−

=

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

( )2

0 43

3 2 8

2 3 8

y x

y x

x y

− = −

= −

− =


Selina Concise Solution Class 10 Chapter 14 Equation of a ...

Documents

Transcript of Selina Concise Solution Class 10 Chapter 14 Equation of a ...