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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