Show that the slopes of the lines represented by $3 x^{2} - 4 x y + y^{2} = 0$ differ by 2.

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Solution

Comparing given equation with $a x^{2} + 2 h x y + b y^{2} = 0$
$a = 3, h = - 2, b = 1$

Now, Let $m_{1}$ and $m_{2}$ be the slope of the lines represented by $3 x^{2} - 4 x y + y^{2} = 0$

$m_{1} + m_{2} = \frac{- 2 h}{b} = \frac{4}{1} = 4.................. (1) m_{1} m_{2} = \frac{a}{b} = - \frac{3}{1} = 3.................. (2)$

Now, $m_{1} - m_{2} = \sqrt{(m_{1} + m_{2})^{2} - 4 m_{1} m_{2}} = \sqrt{4^{2} - 4 \times 3} = 2$

Hence Proved

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