If the equation ax² + 2hxy + by² = 0 represents a pair of distinct (i.e., intersecting) lines, then show that the combined equation of the pair of bisectors of the angles between these lines is $\mathrm{h}\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)=(\mathrm{a}-\mathrm{b})\mathrm{xy}.$

$\text{ Let }{\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2=0\text{ represents the lines }$

$\begin{array}{l}\mathrm{y}={\mathrm{m}}_1\mathrm{x} ...\left(1\right)\\ \mathrm{y}={\mathrm{m}}_2\mathrm{x} ...\left(2\right)\end{array}$
$\text{ Then }{\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{bx}}^2=\left({\mathrm{m}}_1\mathrm{x}-\mathrm{y}\right)\left({\mathrm{m}}_2\mathrm{x}-\mathrm{y}\right)$
on comparing coefficients  on both sides we get
${\mathrm{m}}_1+{\mathrm{m}}_2=\frac{-2\mathrm{h}}{\mathrm{b}},{\mathrm{m}}_1{\mathrm{m}}_2=\frac{\mathrm{a}}{\mathrm{b}}$
Let the one of bisector of the lines (1) & (2) is
y = mx ...(3)
$\mathrm{m}=\frac{\mathrm{y}}{\mathrm{x}}$
Let the lines (1), (2) &(3) makes angles ${\mathrm{\theta }}_1,{\mathrm{\theta }}_2,\mathrm{\theta }$ with x–axis so
${\mathrm{m}}_1=\mathrm{Tan}{\mathrm{\theta }}_1,{\mathrm{m}}_2=\mathrm{Tan}{\theta }_2,\mathrm{m}=\underset{ }{\mathrm{Tan}\mathrm{\theta }}$
From figure we have
$\begin{array}{l}\begin{array}{rl} & \mathrm{\theta }-{\mathrm{\theta }}_1={\mathrm{\theta }}_2-\mathrm{\theta }\\ & \Rightarrow \mathrm{\theta }+\mathrm{\theta }={\mathrm{\theta }}_1+{\mathrm{\theta }}_2\\ & \Rightarrow 2\mathrm{\theta }={\mathrm{\theta }}_1+{\mathrm{\theta }}_2\\ & \text{ Now }\mathrm{Tan}2\mathrm{\theta }=\mathrm{Tan}\left({\mathrm{\theta }}_1+{\mathrm{\theta }}_2\right.)\end{array}\\ \\ \Rightarrow \frac{2\mathrm{Tan}\mathrm{\theta }}{1-{\tan }^2\mathrm{\theta }}=\frac{\mathrm{Tan}{\mathrm{\theta }}_1+\mathrm{Tan}{\mathrm{\theta }}_2}{1-\mathrm{Tan}{\mathrm{\theta }}_1\cdot \mathrm{Tan}{\mathrm{\theta }}_2}\\ \Rightarrow \frac{2\mathrm{m}}{1-{\mathrm{m}}^2}=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{1-{\mathrm{m}}_1{\mathrm{m}}_2}\\ \Rightarrow \frac{2\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}{1-\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}}=\frac{\frac{-2\mathrm{h}}{\mathrm{b}}}{1-\frac{\mathrm{a}}{\mathrm{b}}}\\ \Rightarrow \frac{\frac{2\mathrm{y}}{\mathrm{x}}}{\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{{\mathrm{x}}^2}}=\frac{\frac{-2\mathrm{h}}{\mathrm{b}}}{\frac{\mathrm{b}-\mathrm{a}}{\mathrm{b}}}\\ \Rightarrow \frac{2\mathrm{xy}}{{\mathrm{x}}^2-{\mathrm{y}}^2}=\frac{-2\mathrm{h}}{-(\mathrm{a}-\mathrm{b})}\\ \Rightarrow \frac{\mathrm{xy}}{{\mathrm{x}}^2-{\mathrm{y}}^2}=\frac{-\mathrm{h}}{-(\mathrm{a}-\mathrm{b})}\\ \Rightarrow \mathrm{h}\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)=(\mathrm{a}-\mathrm{b})\mathrm{xy}\end{array}$