If one of the lines given by $6 x^{2} - x y + 4 c y^{2} = 0$ is $3 x + 4 y = 0$ , then $c$ equals

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Solution

The correct option is D -3
The equation $a x^{2} + 2 h x y + b y^{2} = 0$
$= (y - m_{1} x) (y - m_{2} x)$
$\Rightarrow m_{1} + m_{2} = - \frac{2 h}{b} = \frac{1}{4 c}$ .....(1)

$\Rightarrow m_{_{1}} m_{2} = \frac{3}{2 c}$
and $3 x + 4 y = 0 \Rightarrow m_{1} = - \frac{3}{4}$
$∴ m_{2} = - \frac{2}{c}$
Now by $(1)$ we have $- (\frac{3}{4} + \frac{2}{c}) = \frac{1}{4 c}$
$\Rightarrow - \frac{3}{4} = \frac{1}{4 c} + \frac{2}{c}, \frac{3}{4} = \frac{1}{4 c} + \frac{8}{4 c}$ $- \frac{3}{4} = \frac{9}{4 c}$

$∴ c = - 3$

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