Answer
$y-0=\dfrac{1}{2}(x-3)$
Work Step by Step
With the given points, $
(3,0)$ and $(-1,-2),$ then
\begin{align*}
y_1&=
0
,\\y_2&=
-2
,\\x_1&=
3
,\text{ and }\\ x_2&=
-1
.\end{align*}
Using $m=\dfrac{y_1-y_2}{x_1-x_2}$ or the Slope Formula, the slope, $m,$ of the line connecting the two given points is
\begin{align*}
m&=
\dfrac{0-(-2)}{3-(-1)}
\\\\&=
\dfrac{0+2}{3+1}
\\\\&=
\dfrac{2}{4}
\\\\&=
\dfrac{1}{2}
.\end{align*}
Using $
y-y_1=m(x-x_1)
$ or the Point-Slope Form, with $m=
\dfrac{1}{2}
$ and using the point $
(3,0)
$, then
\begin{align*}
y-0&=\dfrac{1}{2}(x-3)
.\end{align*}
A Point-Slope Form of the line with the given conditions is $
y-0=\dfrac{1}{2}(x-3)
.$
