Answer
$\text{a) Slope-Intercept Form: }
y=-\dfrac{3}{5}x-\dfrac{11}{5}
\\\\\text{b) Standard Form: }
3x+5y=-11$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Find the slope of the given equation, $
3x+5y=6
.$ Then use 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}
3x+5y=6
\\\\
5y=-3x+6
\\\\
\dfrac{5y}{5}=\dfrac{-3x}{5}+\dfrac{6}{5}
\\\\
y=-\dfrac{3}{5}x+\dfrac{6}{5}
.\end{array}
Hence, the slope of the given line is $
m=-\dfrac{3}{5}
.$
Since parallel lines have the same slope, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
m=-\dfrac{3}{5}
\\\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{3}{5}
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-2=-\dfrac{3}{5}(x-(-7))
\\\\
y-2=-\dfrac{3}{5}(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{3}{5}(x)-\dfrac{3}{5}(7)
\\\\
y-2=-\dfrac{3}{5}x-\dfrac{21}{5}
\\\\
y=-\dfrac{3}{5}x-\dfrac{21}{5}+2
\\\\
y=-\dfrac{3}{5}x-\dfrac{21}{5}+\dfrac{10}{5}
\\\\
y=-\dfrac{3}{5}x-\dfrac{11}{5}
.\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{3}{5}x-\dfrac{11}{5}
\\\\
5(y)=5\left( -\dfrac{3}{5}x-\dfrac{11}{5} \right)
\\\\
5y=-3x-11
\\\\
3x+5y=-11
.\end{array}
The equation of the line is
\begin{array}{l}\require{cancel}
\text{a) Slope-Intercept Form: }
y=-\dfrac{3}{5}x-\dfrac{11}{5}
\\\\\text{b) Standard Form: }
3x+5y=-11
.\end{array}