Answer
a) $y=\dfrac{2}{3}x-\dfrac{16}{3}$
b) $2x-3y-16=0$
Work Step by Step
a) The equation of a line in slope-intercept form is:
$$y=mx+b,$$
where $m$ is the slope and $b$ in the $y$-intercept.
We are given:
$$\begin{align*}
P(-1,-6)&\text{ belongs to the line}\\
Q(2,-4)&\text{ belongs to the line}.
\end{align*}$$
We calculate the slope $m$ using the coordinates of the two points $P$ and $Q$:
$$\begin{align*}
m&=\dfrac{y_Q-y_P}{x_Q-x_P}\\
&=\dfrac{-4-(-6)}{2-(-1)}\\
&=\dfrac{2}{3}.
\end{align*}$$
We substitute $m=-1/2$ in Eq. $(1)$:
$$y=\dfrac{2}{3}x+b.\tag2$$
Determine $b$ using the coordinates of the point $Q$ in Eq. $(2)$:
$$\begin{align*}
-4&=\dfrac{2}{3}(2)+b\\
-4&=\dfrac{4}{3}+b\\
b&=-4-\dfrac{4}{3}\\
&=-\dfrac{16}{3}
\end{align*}$$
The equation of the line is:
$$y=\dfrac{2}{3}x-\dfrac{16}{3}.$$
b) The equation of a line in general form is:
$$Ax+By+C=0.$$
Rewrite the equation $y=(2/3)x-(16/3).$ in general form:
$$\begin{align*}
3y&=2x-16\\
2x-3y-16&=0.
\end{align*}$$
c) Graph the line using the points $P$ and $Q$:
