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Let L be the line joining the origin to the point of intersection of the lines represented by $2 x^{2} - 3 x y - 2 y^{2} + 10 x + 5 y = 0 .$ If L is perpendicular to the line $k x + y + 3 = 0$ , then k =

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$\frac{- 1}{2}$

$- 1$

$\frac{1}{2}$

$\frac{1}{3}$

answer is B.

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Detailed Solution

$2 x^{2} - 3 x y - 2 y^{2} + 10 x + 5 y = 0 = s$

$\frac{d s}{d x} = 0 \to 4 x - 3 y + 10 = 0$

$\frac{d s}{d x} = 0 \to - 3 x - 4 y + 5 = 0$

Solve $\to x = - 1, y = 2$

$P (x, y) = (- 1, 2) \to slope of \bar{O P} = m_{1} = - 2$

slope of $k x + y + 3 = 0$ is $m_{2} = - k$

$m_{1} m_{2} = - 1 \Rightarrow k = \frac{- 1}{2}$

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