Answer
$\text{a) Slope-Intercept Form: }
y=-\dfrac{1}{2}x-\dfrac{3}{2}
\\\\\text{b) Standard Form: }
x+2y=-3$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Find the slope of the given equation, $
y=2x
.$ Then use the negative reciprocal of this slope and the given point, $
(-7,2)
,$ to find the equation of the needed line. Give the equation in the Slope-Intercept Form and in the Standard Form.
$\bf{\text{Solution Details:}}$
In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the given equation is equivalent to
\begin{array}{l}\require{cancel}
y=2x
.\end{array}
Hence, the slope of the given line is $
m=2
.$
Since perpendicular lines have negative reciprocal slopes, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
m=-\dfrac{1}{2}
\\\text{Through: }
(-7,2)
.\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=-7
,\\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-(-7))
\\\\
y-2=-\dfrac{1}{2}(x+7)
.\end{array}
Using the properties of equality, 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}(7)
\\\\
y-2=-\dfrac{1}{2}x-\dfrac{7}{2}
\\\\
y=-\dfrac{1}{2}x-\dfrac{7}{2}+2
\\\\
y=-\dfrac{1}{2}x-\dfrac{7}{2}+\dfrac{4}{2}
\\\\
y=-\dfrac{1}{2}x-\dfrac{3}{2}
.\end{array}
Using the properties of equality, 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-\dfrac{3}{2}
\\\\
2(y)=2\left( -\dfrac{1}{2}x-\dfrac{3}{2} \right)
\\\\
2y=-x-3
\\\\
x+2y=-3
.\end{array}
The equation of the line is
\begin{array}{l}\require{cancel}
\text{a) Slope-Intercept Form: }
y=-\dfrac{1}{2}x-\dfrac{3}{2}
\\\\\text{b) Standard Form: }
x+2y=-3
.\end{array}
