#### Answer

The equation of the perpendicular line is $\underline{y-3=5\left( x+5 \right);5x-y+28=0}$.

#### Work Step by Step

Consider the equation of the straight line as follows.
$\begin{align}
& x+5y-7=0 \\
& y=\left( -\frac{1}{5} \right)x+\frac{7}{5}
\end{align}$
The slope of this line is ${{m}_{1}}=-\frac{1}{5}$ .
Let the slope of the line perpendicular to this line be ${{m}_{2}}$.
Then,
$\begin{align}
& {{m}_{1}}{{m}_{2}}=-1 \\
& \left( -\frac{1}{5} \right){{m}_{2}}=-1 \\
& {{m}_{2}}=5
\end{align}$
Let $\left( x,y \right)$ be any point on the perpendicular line passing through $\left( -5,3 \right)$.
$\begin{align}
& \frac{y-3}{x+5}=5 \\
& y-3=5\left( x+5 \right)
\end{align}$
Therefore, the equation of the perpendicular line in point-slope form is:
$y-3=5\left( x+5 \right)$
And, in general form it is as follows:
$5x-y+28=0$
Therefore, the equation of the perpendicular line is $y-3=5\left( x+5 \right);5x-y+28=0$.