The equation of perpendicular line on $x+3 y=7$ is $3 x-y+\lambda=0$.
Since, it is passes through $(3,8)$.
$\begin{aligned}
& \therefore \quad 9-8+\lambda=0 \\
& \Rightarrow \quad \lambda=-1 \\
& \therefore \quad 3 x-y-1=0 \\
&
\end{aligned}$
The point of intersection of lines $x+3 y=7$ and $3 x-y-1=0$ is $(1,2)$, which is the foot of a point.
Let the image of a point $(3,8)$ be $\left(x_1, y_1\right)$.
$\begin{array}{ll}
\therefore & \frac{3+x_1}{2}=1, \frac{8+y_1}{2}=2 \\
\Rightarrow & x_1=-1, y_1=-4
\end{array}$