We have, vertices of triangle are $A(1,7), B(-5,1)$ and $C(-1,4)$ Equation of line $A B$ is


$$
\begin{aligned}
& y+1=\frac{7+1}{1+5}(x+5) \Rightarrow y+1=\frac{4}{3}(x+5) \\
& \Rightarrow \quad 4 x-3 y+17=0 \\
&
\end{aligned}
$$
Equation of line $B C$ is
$$
\begin{aligned}
y+1 & =\frac{2+1}{-1+5}(x+5) \\
\Rightarrow 3 x-4 y+11 & =0
\end{aligned}
$$
Equation of bisector of $\angle A B C$ is
$$
\begin{array}{cc}
& \frac{4 x-3 y+17}{5}= \pm \frac{3 x-4 y+11}{5} \\
\Rightarrow & 4 x-3 y+17=3 x-4 y+11 \\
\text { or } & 4 x-3 y+17=-3 x+4 y-11 \\
\Rightarrow \quad & x+y-6=0 \text { or } 7 x-7 y+28=0 \\
\Rightarrow \quad & x+y-6=0 \text { or } x-y+4=0
\end{array}
$$
$\therefore$ Equation are $x-y+4=0$