# Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14E Solutions

## Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14E Solutions

### Exercise 14E

(Q1) Find the value of K for which the lines kx – 5y + 4 = 0 and 5x -2y + 5 = 0 are perpendicular to each other.

Solution:

Given equation of –

Lines are kx – 5y + 4 = 0 ——- (1)

And 5x – 2y + 5 = 0 —– (2)

First we have to convert equation (1) in general form of y = mx + c

Kx – 5y + 4 = 0

-5y = -kx -4

5y = kx + 4

y = kx + 4/5

y = k /5 x + 4/5

∴ Slope (m1) = k/5

Now, we have to convert equation (2) in the form of y = mx + c

5x – 2y + 5 = 0

-2y = – 5x – 5

2y = 5x + 5

y = 5x +5/2

y = 5/2 x + 5/2

∴ Slope (m2) = 5/2

Also, given that, two lines are perpendicular,

We know that,

If two lines are perpendicular then the product of slopes of it two lines is -1.

∴ m1 × m2 = -1

(k/5) × (5/2) = -1

k/2 = -1

k = -2

(Q2) 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 lines are

(k – 2) x + (k + 3) y – 5 = 0 —— (1) is perpendicular to the line

2x – y + 7 = 0 ——- (2)

First we have to convert equation (1) into the general form of y = mx + c

(k-2) x + (k+3)y – 5 = 0

(k + 3) y = – (k-5)x + 5

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

∴ Slope (m1) = – (k-2)/(k+3)

Now, we have to convert equation (2) into the form of y = mx + c.

2x – y +7 = 0

-y = -2x -7

y = 2x +7

∴ Slope (m2) = 2

Also, given that, the two lines are perpendicular.

We know that,

If two lines are perpendicular then the product of slopes of two lines is -1.

∴ m1 × m2 = -1

-(k-2)/(k+3) × 2 = -1

(2-k)/(k+3) × 2 = -1

4 – 2k/(k +3) = -1

4 – 2k = -1 × (k + 3)

4 – 2k = -k -3

-2k + x = -3 -4

-k = -7

K = 7

(ii) If the two lines are parallel then slopes of two lines are same or equal.

Equation (1) is parallel to the equation (2)

∴ m1 = m2

-(k-2)/(k+3) = 2

(2-k)/(k+3) = 2

2-k = 2× (k+3)

2-k = 2k + 6

– k – 2k = 6-2

-3k = 4

K = 4/-3

Here is your solution of Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14E

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Updated: November 25, 2021 — 2:47 pm