If the pair of lines represented by the equation $6 x^{2} + 17 x y + 12 y^{2} + 22 x + 31 y + 20 = 0$ be $2 x + 3 y + p = 0$ and $3 x + 4 y + q = 0$ , then

p + q = 9

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p^{2} + q^{2} = 0

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3 p + 2 q = 22

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4 p + 3 q = 31

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Solution

The correct options are
A $3 p + 2 q = 22$
B $p + q = 9$
C $4 p + 3 q = 31$
$(2 x + 3 y + p) (3 x + 4 y + q) = 0$
$6 x^{2} + 8 x y + 2 q x + 9 x y + 12 y^{2} + 3 q y + 3 p x + 4 p y + p q = 0$
$6 x^{2} + 17 x y + 12 y^{2} + x (2 q + 3 p) + y (3 q + 4 p) + p q = 0$
$\to 6 x^{2} + 17 x y + 12 y^{2} + 22 x + 31 y + 20 = 0$
Hence comparison gives us
$p q = 20$
$3 p + 2 q = 22$
$4 p + 3 q = 31$
$(4 p + 3 q) - (3 p + 2 q) = 31 - 22$
$p + q = 9$ .
Therefore
$p^{2} + q^{2} = (p + q)^{2} - 2 p q$
$= 81 - 2 (20)$
$= 81 - 40$
$= 41$

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