#### Answer

$\text{a) Slope-Intercept Form: }
y=2x-10
\\\text{b) Standard Form: }
2x-y=10$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the Slope Formula t of find the slope of the line passing through $(3,7)$ and $(5,6).$ Then use the Point-Slope Form of linear equations with the given point, $
(4,-2)
$ to find the equation of the needed line. Finally, Express the answer in both the Slope-Intercept and Standard forms.
$\bf{\text{Solution Details:}}$
With the given points,
\begin{array}{l}\require{cancel}
y_1=
7
,\\y_2=
6
,\\x_1=
3
,\text{ and }\\ x_2=
5
.\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{7-6}{3-5}
\\\\
m=\dfrac{1}{-2}
\\\\
m=-\dfrac{1}{2}
.\end{array}
Since perpendicular lines have negative reciprocal slopes, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
m=2
\\\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=
2
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-(-2)=2(x-4)
\\\\
y+2=2(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=2(x-4)
\\\\
y+2=2(x)+2(-4)
\\\\
y+2=2x-8
\\\\
y=2x-8-2
\\\\
y=2x-10
.\end{array}
In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=2x-10
\\\\
-2x+y=-10
\\\\
-1(-2x+y)=-1(-10)
\\\\
2x-y=10
.\end{array}
Hence, $
\text{a) Slope-Intercept Form: }
y=2x-10
\\\text{b) Standard Form: }
2x-y=10
.$