Answer
The equation of the line passes through $\left( -3,\ 2 \right)$ and the slope of the line is -6; the point-slope and slope-intercept form is $y+6x+16=0$ and $y=-6x-16$.
Work Step by Step
The slope and a point on the line is given as $\left( -3,\ 2 \right)$ and -6.
Point-Slope form:
Therefore, the equation of the line by point-slope form will be:
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
Substituting the given data in the above equation to get the desired equation of the line:
$\begin{align}
& y-2=-6\left\{ x-\left( -3 \right) \right\} \\
& y-2=-6\left( x+3 \right) \\
& y-2=-6x-18 \\
& y+6x+16=0
\end{align}$
Slope-Intercept form:
Now, the above equation can be rewritten as:
$y=-6x-16$
Comparing the above equation with the general equation of slope-intercept form:
$\begin{align}
& y=mx+c \\
& \text{Therefore,} \\
& m=-6 \\
& c=-16.
\end{align}$
Therefore, the equation of the line passes through $\left( -3,\ 2 \right)$ and the slope of the line is -6; the point-slope and slope-intercept form is $y+6x+16=0$ and $y=-6x-16$.
