#### Answer

The equation of the line that passes through the point $\left( -5,6 \right)$ and is perpendicular to the line that has an x-intercept of $3$ and a y-intercept of $-9$ in slope–intercept form is $y=-\frac{1}{3}x+\frac{13}{3}$.

#### Work Step by Step

First find the equation of the line with an x-intercept of $3$ and a y-intercept of $-9$. This line will pass through $\left( 3,0 \right)\text{ and }\left( 0,-9 \right)$.
Use these points to find the slope as:
$m=\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}$
Put the value of ${{x}_{1}}=3\text{ and }{{x}_{2}}=0$ in the above formula to get:
$\begin{align}
& m=\frac{f\left( 3 \right)-f\left( 0 \right)}{3-0} \\
& =\frac{0-\left( -9 \right)}{3-0} \\
& =\frac{9}{3} \\
& =3
\end{align}$
The slope of the perpendicular line is the negative reciprocal of $3$.
Thus, the slope of the perpendicular line is $-\frac{1}{3}$.
Use the point-slope formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
Substitute the value of the slope of the line $m=-\frac{1}{3}$ and point $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -5,6 \right)$ in equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
The obtained expression is:
$\begin{align}
& y-6=-\frac{1}{3}\cdot \left( x-\left( -5 \right) \right) \\
& y-6=-\frac{1}{3}\cdot \left( x+5 \right)
\end{align}$
Use the distributive property to get:
$\begin{align}
& y-6=-\frac{1}{3}x-\frac{5}{3} \\
& y=-\frac{1}{3}x-\frac{5}{3}+6 \\
& y=-\frac{1}{3}x+\frac{13}{3}
\end{align}$
Thus, the required equation of the line is $y=-\frac{1}{3}x+\frac{13}{3}$.