Answer
$\text{a) Slope-Intercept Form: }
y=\dfrac{2}{3}x+\dfrac{14}{3}
\\\text{b) Standard Form: }
2x-3y=-14$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the Slope Formula t of find the slope of the line passing through $(3,9)$ and $(6,11).$ Then use the Point-Slope Form of linear equations with the given point, $
(-2,5)
$ to find the equation of the needed line. Finally, Express the answer in both the Slope-Intercept and Standard forms.
$\bf{\text{Solution Details:}}$
With the given points,
\begin{array}{l}\require{cancel}
y_1=
9
,\\y_2=
11
,\\x_1=
3
,\text{ and }\\ x_2=
6
.\end{array}
Using $m=\dfrac{y_1-y_2}{x_1-x_2}$ or the Slope Formula, the slope, $m,$ of the line is
\begin{array}{l}\require{cancel}
m=\dfrac{y_1-y_2}{x_1-x_2}
\\\\
m=\dfrac{9-11}{3-6}
\\\\
m=\dfrac{-2}{-3}
\\\\
m=\dfrac{2}{3}
.\end{array}
Since parallel lines have same slopes, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
m=\dfrac{2}{3}
\\\text{Through: }
(-4,2)
.\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=
-4
,\\m=
\dfrac{2}{3}
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-2=\dfrac{2}{3}(x-(-4))
\\\\
y-2=\dfrac{2}{3}(x+4)
.\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{2}{3}(x+4)
\\\\
y-2=\dfrac{2}{3}(x)+\dfrac{2}{3}(4)
\\\\
y-2=\dfrac{2}{3}x+\dfrac{8}{3}
\\\\
y=\dfrac{2}{3}x+\dfrac{8}{3}+2
\\\\
y=\dfrac{2}{3}x+\dfrac{8}{3}+\dfrac{6}{3}
\\\\
y=\dfrac{2}{3}x+\dfrac{14}{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{2}{3}x+\dfrac{14}{3} \right)
\\\\
3y=2x+14
\\\\
-2x+3y=14
\\\\
-1(-2x+3y)=-1(14)
\\\\
2x-3y=-14
.\end{array}
Hence, $
\text{a) Slope-Intercept Form: }
y=\dfrac{2}{3}x+\dfrac{14}{3}
\\\text{b) Standard Form: }
2x-3y=-14
.$
