Answer
The equation of the line that passes through the point $\left( -6,4 \right)$ and is perpendicular to the line that has an x-intercept of $2$ and a y-intercept of 4 in slope–intercept form is $y=-\frac{1}{2}x+1$.
Work Step by Step
First find the equation of the line with x-intercept of $2$ and a y-intercept of 4. Then, this line will pass through $\left( 2,0 \right)\text{ and }\left( 0,-4 \right)$.
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 ${{x}_{1}}=2\text{ and }{{x}_{2}}=0$ in the above formula to get:
$\begin{align}
& m=\frac{f\left( 2 \right)-f\left( 0 \right)}{2-0} \\
& =\frac{0-\left( -4 \right)}{2-0} \\
& =\frac{4}{2} \\
& =2
\end{align}$
The slope of the perpendicular line is the negative reciprocal of $2$.
Thus, the slope of the perpendicular line is $-\frac{1}{2}$.
Use the point-slope formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
Substitute the value of the slope of the line $m=-\frac{1}{2}$ and point $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -6,4 \right)$ in equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
The obtained expression is:
$\begin{align}
& y-4=-\frac{1}{2}\cdot \left( x-\left( -6 \right) \right) \\
& y-4=-\frac{1}{2}\cdot \left( x+6 \right)
\end{align}$
Use the distributive property to get:
$\begin{align}
& y-4=-\frac{1}{2}x-3 \\
& y=-\frac{1}{2}x+1
\end{align}$
Thus, the required equation of the line is $y=-\frac{1}{2}x+1$.
