Answer
$\text{a) Slope-Intercept Form: }
y=\dfrac{1}{3}x+\dfrac{10}{3}
\\\text{b) Standard Form: }
x-3y=-10
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Find the slope of the line defined by the given equation, $
-x+3y=12
.$ Then use the Point-Slope Form of linear equations with the given point, $
(-1,3)
$ 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+3y=12
\\\\
3y=x+12
\\\\
\dfrac{3y}{3}=\dfrac{x}{3}+\dfrac{12}{3}
\\\\
y=\dfrac{1}{3}x+4
.\end{array}
Using $y=mx+b$ or the Slope-Intercept Form, where $m$ is the slope, then the slope of the line with the equation above is
\begin{array}{l}\require{cancel}
m=\dfrac{1}{3}
.\end{array}
Since parallel lines have the same slope, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
\text{Through: }
(-1,3)
\\
m=\dfrac{1}{3}
.\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=3
,\\x_1=
-1
,\\m=
\dfrac{1}{3}
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-3=\dfrac{1}{3}(x-(-1))
\\\\
y-3=\dfrac{1}{3}(x+1)
.\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}{3}(x)+\dfrac{1}{3}(1)
\\\\
y-3=\dfrac{1}{3}x+\dfrac{1}{3}
\\\\
y=\dfrac{1}{3}x+\dfrac{1}{3}+3
\\\\
y=\dfrac{1}{3}x+\dfrac{1}{3}+\dfrac{9}{3}
\\\\
y=\dfrac{1}{3}x+\dfrac{10}{3}
.\end{array}
In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
3(y)=3\left( \dfrac{1}{3}x+\dfrac{10}{3} \right)
\\\\
3y=x+10
\\\\
-x+3y=10
\\\\
-1(-x+3y)=-1(10)
\\\\
x-3y=-10
.\end{array}
Hence, $
\text{a) Slope-Intercept Form: }
y=\dfrac{1}{3}x+\dfrac{10}{3}
\\\text{b) Standard Form: }
x-3y=-10
.$