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Question
Find the equation of the lines passing through the point of intersection lines 4x − y + 3 = 0 and 5x + 2y + 7 = 0, and perpendicular to x − 2y + 1 = 0
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Solution
The equation of the straight line passing through the point of intersection of the lines.
4x – y + 3 = 0 and 5x + 2y + 7 = 0 is
(4x – y + 3) + λ(5x + 2y + 7) = 0 ......(1)
Perpendicular to x – 2y + 1 = 0
Given that the line (1) perpendicular to the line
x – 2y + 1 = 0 ......(3)
(1) ⇒ (4x – y + 3) + λ(5x + 2y + 7) = 0
4x – y + 3 + 5λx + 2λy + 7λ = 0
(4 + 5λ)x + (2λ – 1 )y + (3 + 7λ) = 0 ......(4)
Slope of this line (3) = `(4 + 5lambda)/(2lambda - 1)`
Slope of line (2) = `- 1/(-2) = 1/2`
Given that line (3) and line (4) are perpendicular
∴ `- (4 + 5lambda)/(2lambda - 1) xx 1/2` = – 1
`(4 + 5lambda)/(2(2lambda - 1))` = 1
4 + 5λ = 2(2λ – 1)
4 + 5λ = 4λ – 2
λ = – 6
Substituting the value of λ in equation (1) we have
(4x – y + 3) – 6(5x + 2y + 7) = 0
4x – y + 3 – 30x – 12y – 42 = 0
– 26x – 13y – 39 =0
2x + y + 3 = 0
which is the required equation.
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