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If $4 x^{2} + 2 l x y + y^{2} + 6 x + k y - 10 = 0$ represents a pair of parallel lines then

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k=-22,l=-12

k=-16,l=9

k=-3,l=2

$k = \pm 3, l = \pm 2$

answer is A.

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Detailed Solution

$l^{2} = 4 \Rightarrow l = \pm 2$

$| \begin{array}{ccc} 4 & \pm 2 & 3 \\ \pm 2 & 1 & \frac{k}{2} \\ 3 & \frac{k}{2} & - 10 \end{array} | = 0$

$\Rightarrow 4 (- 10 - \frac{k^{2}}{4}) \pm 2 (\pm 20 - \frac{3 k}{2}) + 3 (\pm k - 3) = 0$

$\Rightarrow k^{2} \pm 6 k + 9 = 0 \Rightarrow {(k \pm 3)}^{2} = 0 \Rightarrow k = \pm 3$

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