#### Answer

$ y+6=-3\left( x+2 \right)$ ; $ y=-3x-12$

#### Work Step by Step

The equation of the line to which the line is perpendicular in the slope-intercept form will be:
$\begin{align}
& 0=x-3y+9 \\
& -3y=x-9 \\
& y=\frac{1}{3}x+3
\end{align}$
The slope of this line is $\frac{1}{3}$; thus the slope of any line perpendicular to this line will be -3; using $ m=-3$ and the point $\left( -2,\ -6 \right)$ we will rewrite the equation in point-slope form:
$\begin{align}
& y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\
& y-\left( -6 \right)=-3\left( x-\left( -2 \right) \right) \\
& y+6=-3\left( x+2 \right)
\end{align}$
Slope-intercept form:
$\begin{align}
& y+6=-3x-6 \\
& y=-3x-12
\end{align}$
Thus, the equation of the line in point-slope form is $ y+6=-3\left( x+2 \right)$ and the equation of the line in slope-intercept form is $ y=-3x-12$, which passes through point $\left( -2,\ -6 \right)$ and is perpendicular to the line whose equation is $ x-3y+9=0$.