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Joint equation of the diagonal of the square formed by the pair of lines $x y + 4 x - 3 y - 12 = 0$ and $x y - 3 x + 4 y - 12 = 0$ is

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$x^{2} - y^{2} + x - y = 0$

$x^{2} - y^{2} + x + y = 0$

$x^{2} + 2 x y + y^{2} + x + y = 0$

$x^{2} - 2 x + y^{2} + x - y = 0$

answer is A.

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Detailed Solution

$a = 4, b = - 3, c = - 3, d = 4$

equation of diagonals are $(a - c) x + (b - d) y + a b - c d = 0$

$\Rightarrow x - y = 0$

$(a - c) x - (b - d) y + a d - b c = 0$

$\Rightarrow x + y + 1 = 0$

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