Question

Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0

Answer 1

Given lines are

                                     5x − 6y = 2        ...(1)
                                     3x + 2y = 10      ...(2)
(1) × 1 ⇒                      5x − 6y =  2       ...(3)
(2) × 3 ⇒                      9x + 6y = 30      ...(4)
By adding (3) and (4) ⇒ 14x     = 32

x = `32/14= 16/7`

Substitute the value of x = `16/7` in (2)

`3 xx 16/7 + 2y` = 10

⇒ 2y = `10 - 48/7`

2y = `(70 - 48)/7`

⇒ 2y = `22/7`

y =`22/(2 xx 7) = 11/7`

The point of intersect is `(16/7, 11/7)`

Equation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0

This line passes through `(16/7, 11/7)`

`7(16/7) + 4(11/7) + "k"` = 0

⇒ `16 + 44/7 + "k"` = 0

`(112 + 44)/7 + "k"` = 0

⇒ `156/7 + "k"` = 0

k = `-156/7`

Equation of the line is `7x + 4y - 156/7` = 0

49x + 28y – 156 = 0