Answer
$\text{a) Slope-Intercept Form: }
y=\dfrac{1}{2}x-1
\\\text{b) Standard Form: }
x-2y=2
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Find the slope of the line defined by the given equation, $
-x+2y=10
.$ Then use the Point-Slope Form of linear equations with the given point, $
(-2,-2)
$ to find the equation of the needed line. Finally, Express the answer in both the Slope-Intercept and Standard forms.
$\bf{\text{Solution Details:}}$
In the form $y=mx+b,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
-x+2y=10
\\\\
2y=x+10
\\\\
\dfrac{2y}{2}=\dfrac{x}{2}+\dfrac{10}{2}
\\\\
y=\dfrac{1}{2}x+5
.\end{array}
Using $y=mx+b$ or the Slope-Intercept Form, where $m$ is the slope, then the slope of the equation above is
\begin{array}{l}\require{cancel}
m=\dfrac{1}{2}
.\end{array}
Since parallel lines have the same slope, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
\text{Through: }
(-2,-2)
\\
m=\dfrac{1}{2}
.\end{array}
Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions,
\begin{array}{l}\require{cancel}
y_1=-2
,\\x_1=
-2
,\\m=
\dfrac{1}{2}
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-(-2)=\dfrac{1}{2}(x-(-2))
\\\\
y+2=\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+2=\dfrac{1}{2}(x)+\dfrac{1}{2}(2)
\\\\
y+2=\dfrac{1}{2}(x)+1
\\\\
y=\dfrac{1}{2}x+1-2
\\\\
y=\dfrac{1}{2}x-1
.\end{array}
In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
2(y)=2\left( \dfrac{1}{2}(x)-1 \right)
\\\\
2y=x-2
\\\\
-x+2y=-2
\\\\
-1(-x+2y)=-1(-2)
\\\\
x-2y=2
.\end{array}
Hence, $
\text{a) Slope-Intercept Form: }
y=\dfrac{1}{2}x-1
\\\text{b) Standard Form: }
x-2y=2
.$
