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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
Solution
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
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