lf $a x^{2} + 6 x y + b y^{2} - 10 x + 10 y - 6 = 0$ represents a pair of perpendicular straight lines, then $| a |$ is equal to

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Solution

The correct option is B $4$
$a x^{2} + 6 x y + b y^{2} - 10 x + 10 y - 6 = 0$ represent pair of lines when
$Δ = 0 \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} a & 3 & - 5 3 & b & 5 - 5 & 5 & - 6 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow a (- 6 b - 25) - 3 (- 18 + 25) - 5 (15 + 5 b) = 0 \Rightarrow - 6 a b - 25 a - 21 - 75 - 25 b = 0$
$\Rightarrow 25 a + 25 b + 6 a b + 96 = 0$ ...(1)
And represent pair of perpendicular lines when
$a + b = 0$ ...(2)
Solving (1) and (2)
$- 6 a^{2} + 96 = 0 \Rightarrow a^{2} = 16 \Rightarrow | a | = 4$

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