If $6 x^{2} - 5 x y + b y^{2} + 4 x + 7 y + c = 0$ represents a pair of perpendicular lines, then :

b = 6, c = - 2

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b = - 6, c = 2

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b = - 6, c = - 2

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b = - 2, c = - 6

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Solution

The correct option is A $b = 6, c = - 2$
$6 x^{2} - 5 x y + b y^{2} + 4 x + 7 y + c = 0 - (1)$
Eq $(1)$ respresents a pair of perpendicular lines
then $b + 6 = 0$
$b = - 6$
$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 6 & - \frac{5}{2} & 2 - \frac{5}{2} & - 6 & 7 / 2 2 & 7 / 2 & c \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
$c = - 2$

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