The difference of the slopes of the lines $3 x^{2} - 4 x y + y^{2} = 0$ is

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Solution

The correct option is B 2
Consider the given pair of equation.

$3 x^{2} - 4 x y + y^{2} = 0$

We know that the standard joint equation of two line is,

$a x^{2} + 2 h x y + b y^{2} = 0$

Therefore,

$a = 3$

$2 h = - 4$

$b = 1$

Let $m_{1}$ and $m_{2}$ be the slopes of both the lines. Then,

$m_{1} + m_{2} = - \frac{2 h}{b}$

$m_{1} + m_{2} = - \frac{(- 4)}{1} = 4$

And,

$m_{1} m_{2} = \frac{a}{b}$

$m_{1} m_{2} = \frac{3}{1} = 3$

We know that,

${(m_{1} - m_{2})}^{2} = {(m_{1} + m_{2})}^{2} - 4 m_{1} m_{2}$

${(m_{1} - m_{2})}^{2} = {(4)}^{2} - 4 (3)$

${(m_{1} - m_{2})}^{2} = 16 - 12$

${(m_{1} - m_{2})}^{2} = 4$

$| m_{1} - m_{2} | = 2$

Hence, the difference of the slopes is $2$ .

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