Equation of the perpendicular bisector of the sides $A B$ and $A C$ of $\triangle A B C$ are $x-y+5=0$ and $x+2 y=0$



$B$ is the image of $A$ with respect to line
$$
\begin{aligned}
& \quad x-y+5=0 \\
& =\frac{x_2-1}{1}=\frac{y_2+2}{-1}=-2\left(\frac{1+2+5}{2}\right) \\
& =\frac{x_2-1}{1}=\frac{y_2+2}{-1}=-8 \\
& \therefore \quad x_2=-7, y_2=6 \\
& B(-7,6) \text { is image of } A(1,-2) \text { with respect to line } \\
& C\left(x_3, y_3\right) \text { is } \\
& x+2 y=0 \quad \frac{x_3-1}{1}=\frac{y_3+2}{2}=\frac{-2(1-4)}{5} \\
& \therefore \quad x_3=\frac{11}{5}, y_3=\frac{2}{5}
\end{aligned}
$$
Solving $x-y+5=0$
and
$$
x+2 y=0
$$

We get $x=-\frac{10}{3}, y=\frac{5}{3}$
Coordinate of $D=\left(\frac{-7+\frac{11}{5}}{2}, \frac{6+\frac{2}{5}}{2}=\frac{-12}{5}, \frac{16}{5}\right)$
Equation of perpendicular bisector of side $B C$
$$
y-\frac{5}{3}=\frac{48-25}{-36+50}\left(x+\frac{10}{3}\right) \Rightarrow 23 x-14 y+100=0
$$