# Show that the line y=mx bisects the angle between lines ax^2-2hxy+by^2=0 if h(1-m^2)+m(a-b)=0?

Aug 6, 2018

Let the equations $y - {m}_{1} x = 0 \mathmr{and} y - {m}_{2} x = 0$ are two straight lines represented by the given equation of pair of straight lines.Here ${m}_{1} = \tan \alpha \mathmr{and} {m}_{2} = \tan \beta \mathmr{and} \beta > \alpha$

Hence

$\left(y - {m}_{1}\right) \left(y - {m}_{2} x\right) = {y}^{2} - \frac{2 h}{b} x y + \frac{a}{b} {x}^{2}$

So

${m}_{1} + {m}_{2} = = \frac{2 h}{b} \mathmr{and} {m}_{1} {m}_{2} = \frac{a}{b}$

If $\theta$ be the angle subtended by angle bisector ($y = m x$) of the pair of straight line with the positive direction of X-axis ,then $m = \tan \theta$

Now it is obvious that

$\theta - \alpha = \beta - \theta$

So

$\alpha + \beta = 2 \theta$

$\implies \tan \left(\alpha + \beta\right) = \tan \left(2 \theta\right)$

$\implies \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$

$\implies \frac{{m}_{1} + {m}_{2}}{1 - {m}_{1} {m}_{2}} = \frac{2 m}{1 - {m}^{2}}$

$\implies \frac{\frac{2 h}{b}}{1 - \frac{a}{b}} = \frac{2 m}{1 - {m}^{2}}$

$\implies \frac{h}{b - a} = \frac{m}{1 - {m}^{2}}$

$\implies h \left(1 - {m}^{2}\right) = \left(b - a\right) m$

$\implies h \left(1 - {m}^{2}\right) + \left(a - b\right) m = 0$