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If $p x^{2} - y^{2} + 3 x + 11 y + q = 0$ represents a pair of perpendicular lines then (p,q)=

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$(1, - 28)$

$(- 1, 28)$

$(- 1, - 14)$

$(- 1, - 28)$

answer is C.

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Detailed Solution

$a = p, 2 h = 0, b = - 1, 2 g = 3, 1 f = 11, c = q$

$a + b = 0 \Rightarrow p = 1$

$a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0 \Rightarrow q = - 28$

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