Example 1. Find the equation of the angle bisector of the two lines $l_{1}: \sqrt{3} x+y-3=0$ and $l_{2}$ : $4 x-3 y-4=0$.
The equation of the family of lines passing through the intersection of $l_{1}, l_{2}$ is
$$
\begin{array}{l} 
\sqrt{3} x+y-3+2(4 x \\
-3 y-4)=0 .
\end{array}
$$

Let $M_{0}\left(0 x, y_{0}\right)$ be any point on the angle bisector of $l_{1}, l_{2}$ other than the intersection point, then the above equation becomes
$$
\begin{array}{l}
\sqrt{3} x_{0}+y_{0}-3+\lambda\left(4 x_{0}-3 y_{0}-4\right)=0, \\
\sqrt{3} x_{0}+y_{0}-3=-\lambda\left(4 x_{0}-3 y_{0}-4\right), \\
\frac{\sqrt{3} x_{0}+y_{0}-3}{\sqrt{(\sqrt{3})^{2}+1^{2}}}=-\frac{4 x_{0}-3 y_{0}-4}{\sqrt{4^{2}+(-3)^{2}}} \\
\times \frac{5}{2} \lambda .
\end{array}
$$

That is
$$
\begin{array}{l}
\sqrt{3} x_{0}+y_{0}-3=-\lambda\left(4 x_{0}-3 y_{0}-4\right) \\
\frac{\sqrt{3} x_{0}+y_{0}-3}{\sqrt{(\sqrt{3})^{2}+1^{2}}}=-\frac{4 x_{0}-3 y_{0}-4}{\sqrt{4^{2}+(-3)^{2}}}
\end{array}
$$

It is only necessary to let $\frac{5}{2} \lambda= \pm 1, \lambda= \pm \frac{2}{5}$. The angle bisectors of $l_{1}, l_{2}$ are:
$$
\begin{array}{l}
(5 \sqrt{3}+8) x_{0}-y_{0}-23=0 \text { and } \\
(5 \sqrt{3}-3) x_{0}+11 y_{0} \\
-7=0
\end{array}
$$