If the equation $12 x^{2} + 7 x y - p y^{2} - 18 x + q y + 6 = 0$ represents two perpendicular lines, then the value of p and q are?

12, 1

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12, - 1

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12, \frac{23}{2}

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12, - \frac{23}{2}

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Solution

The correct option is A $12, 1$
Given equation is $12 x^{2} + 7 x y - p y^{2} - 18 x + 9 q + 6 = 0$
We know that for it to represent a pair of $⊥$ lines, coeff. of $x^{2}$ + coeff. of $y^{2} = 0$
$\Rightarrow 12 - p = 0 \Rightarrow p = 12$
And also, $a = 12, h = \frac{7}{2}, b = - p = - 12, g = - 9, f = \frac{q}{2}$ , $a b c + 2 f g h - b g^{2} - a f^{2} - c h^{2} = 0 \Rightarrow q = 1$ or $q = - \frac{23}{2}$

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