Find the equation of angle bisectors of the pair of lines given by:
x²+4xy+y²=0.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{For} {\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2=0, \mathrm{equation} \mathrm{of} \mathrm{bisector} \mathrm{is} \\ \frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{\mathrm{a}-\mathrm{b}}=\frac{\mathrm{xy}}{\mathrm{h}} \\ \mathrm{Hence} \\ {\mathrm{x}}^2+4\mathrm{xy}+{\mathrm{y}}^2=0 \\ \mathrm{a}=1 , \mathrm{b}=1 \mathrm{and} \mathrm{h}=2 \\ \Rightarrow \frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{1-1}=\frac{\mathrm{xy}}{2} \\ \Rightarrow \frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{0}=\frac{\mathrm{xy}}{2} \\ \Rightarrow {\mathrm{x}}^2-{\mathrm{y}}^2=0 \\ \Rightarrow \left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{x}+\mathrm{y}\right)=0 \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0} \end{gathered}$$

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