Answer
Point-slope equation $ y-6=-3(x+2)$.
Slope-intercept form $ y=-3x$.
Work Step by Step
If the line passes through a point $(x_1,y_1)$ and slope is $m$ then the point-slope form of the perpendicular line's equation is.
$\Rightarrow y-y_1=m(x-x_1)$
From the question we have
$\Rightarrow (x_1,y_1)=(-2,6)$
The equation of the line is
$\Rightarrow y=\frac{1}{3}x+4$
It is in the form $y=mx+c$.
Where slope $m_1=\frac{1}{3}$.
perpendicular lines have slopes negative reciprocal to each other.
Slope of the required line is $m_2=-\frac{1}{m_1}$.
Plug the value of $m_1$.
$\Rightarrow m_2=-\frac{1}{\frac{1}{3}}$
$\Rightarrow m_2=-3$
Substitute all values into the point-slope equation.
$\Rightarrow y-(6)=(-3)(x-(-2))$
Simplify.
$\Rightarrow y-6=-3(x+2)$
The above equation is the point-slope form.
Now isolate $y$.
$\Rightarrow y-6=-3(x+2)$
Use distributive property.
$\Rightarrow y-6=-3x-6$
Add $6$ to both sides.
$\Rightarrow y-6+6=-3x-6+6$
Simplify.
$\Rightarrow y=-3x$
The above equation is the slope-intercept form.
