Answer
$\text{a) Slope-Intercept Form: }
y=-\dfrac{1}{2}x+2
\\\\\text{b) Standard Form: }
x+2y=4$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the equation of the line with the given characeristics:
\begin{array}{l}\require{cancel}
\text{through }
(-2,3) \text{ and } (6,-1)
\end{array}
use the Two-Point Form of linear equations. Express the answer in Slope-Intercept Form and in the Standard Form.
$\bf{\text{Solution Details:}}$
Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-Point Form of linear equations, where
\begin{array}{l}\require{cancel}
x_1=-2
,\\x_2=6
,\\y_1=3
,\\y_2=-1
,\end{array}
the equation of the line is
\begin{array}{l}\require{cancel}
y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)
\\\\
y-3=\dfrac{3-(-1)}{-2-6}(x-(-2))
\\\\
y-3=\dfrac{3+1}{-2-6}(x+2)
\\\\
y-3=\dfrac{4}{-8}(x+2)
\\\\
y-3=-\dfrac{1}{2}(x+2)
.\end{array}
In the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y-3=-\dfrac{1}{2}(x+2)
\\\\
y-3=-\dfrac{1}{2}(x)-\dfrac{1}{2}(2)
\\\\
y-3=-\dfrac{1}{2}x-1
\\\\
y=-\dfrac{1}{2}x-1+3
\\\\
y=-\dfrac{1}{2}x+2
.\end{array}
In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=-\dfrac{1}{2}x+2
\\\\
2(y)=2\left( -\dfrac{1}{2}x+2 \right)
\\\\
2y=-x+4
\\\\
x+2y=4
.\end{array}
Hence, the equation of the line is
\begin{array}{l}\require{cancel}
\text{a) Slope-Intercept Form: }
y=-\dfrac{1}{2}x+2
\\\\\text{b) Standard Form: }
x+2y=4
.\end{array}
