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If ${ax}^{2} + 6 xy + {by}^{2} - 10 x + 10 y - 6 = 0$ represents a pair of perpendicular lines then |a| =

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answer is D.

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Detailed Solution

$a + b = 0 \Rightarrow b = - a Δ \equiv abc + 2 fgh - {af}^{2} - {bg}^{2} - {ch}^{2} = 0 a (- a) (- 6) + 2 (5) (- 5) (3) - a (25) - (- a) (25) - (- 6) (9) = 0 6 a^{2} - 150 + 54 = 0 \Rightarrow a^{2} = 16 \Rightarrow | a | = 4$

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