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Equation of pair of lines passing through origin and making an angle ${Tan}^{- 1} 2$ with the line $4 x - 3 y + 7 = 0$ is

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$4 {(4 x - 3 y)}^{2} - {(3 x + 4 y)}^{2} = 0$

${(4 x - 3 y)}^{2} - {(3 x + 4 y)}^{2} = 0$

${(4 x - 3 y)}^{2} - 3 {(3 x + 4 y)}^{2} = 0$

${(4 x - 3 y)}^{2} - 4 {(3 x + 4 y)}^{2} = 0$

answer is A.

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Detailed Solution

Let $α = T a n^{- 1} (2) \Rightarrow T a n α = 2$

and given line $4 x - 3 y + 7 = 0$

$l = 4, m = - 3, n = 7$

Required Equation is ${(l x + m y)}^{2} - T a n^{2} α {(m x - l y)}^{2} = 0 \Rightarrow {(4 x - 3 y)}^{2} - 4 {(3 x + 4 y)}^{2} = 0$

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