Answer
The equation of the line passing through $\left( -3,\ 6 \right)$ and perpendicular to the line with equation $y=\frac{1}{3}x+4$ in point-slope and slope-intercept form is $y+3x+3=0$ and $y=-3x-3$ respectively.
Work Step by Step
The point on the line is given as $\left( -3,\ 6 \right)$ and the equation of the line parallel to the required line is $y=\frac{1}{3}x+4$.
It is known that if two lines are perpendicular, then the product of their slopes is $-1$.
The slope of the given line equation is $\frac{1}{3}$.
Let the slope of the required line equation be m. Therefore,
$\begin{align}
& \frac{1}{3}m=-1 \\
& m=-3.
\end{align}$
Therefore, the slope of the required line equation will be $-3$.
Point-Slope form:
We know that equation of the line by point-slope form will be:
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
Now, substitute the given data in the above equation to get the desired equation of the line:
$\begin{align}
& y-6=-3\left\{ x-\left( -3 \right) \right\} \\
& y-6=-3\left( x+3 \right) \\
& y-6=-3x-9 \\
& y+3x+3=0
\end{align}$
Slope-Intercept form:
The above equation can be rewritten as:
$y=-3x-3$
Now, compare the above equation with the general equation of slope-intercept form:
$\begin{align}
& y=mx+c. \\
& \text{Therefore,} \\
& m=-3 \\
& c=-3.
\end{align}$
Therefore, the equation of the line passing through $\left( -3,\ 6 \right)$ and perpendicular to the line with equation $y=\frac{1}{3}x+4$ in point-slope and slope-intercept form is $y+3x+3=0$ and $y=-3x-3$ respectively.
