Answer
$\text{slope-intercept form: }
y=-\dfrac{5}{2}x+2
\\\\
\text{standard form: }
5x+2y=4$
Work Step by Step
Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the two-point form of linear equations, the equation of the line passing through $\left(
4,-8
\right)$ and $\left(
-4,12
\right)$ is
\begin{array}{l}\require{cancel}
y-(-8)=\dfrac{-8-12}{4-(-4)}\left( x-4 \right)
\\\\
y+8=\dfrac{-8-12}{4+4}\left( x-4 \right)
\\\\
y+8=\dfrac{-20}{8}\left( x-4 \right)
\\\\
y+8=-\dfrac{5}{2}\left( x-4 \right)
.\end{array}
In the form $y=mx+b$, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y+8=-\dfrac{5}{2}\left( x-4 \right)
\\\\
y+8=-\dfrac{5}{2}x+10
\\\\
y=-\dfrac{5}{2}x+10-8
\\\\
y=-\dfrac{5}{2}x+2
.\end{array}
In the form $Ax+By=C$, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=-\dfrac{5}{2}x+2
\\\\
2(y)=2\left( -\dfrac{5}{2}x+2 \right)
\\\\
2y=-5x+4
\\\\
5x+2y=4
.\end{array}
Hence, the different forms of the equation of the line with the given conditions are
\begin{array}{l}\require{cancel}
\text{slope-intercept form: }
y=-\dfrac{5}{2}x+2
\\\\
\text{standard form: }
5x+2y=4
.\end{array}