Answer
$\text{a) Slope-Intercept Form: }
y=\dfrac{7}{5}x+\dfrac{16}{5}
\\\\\text{b) Standard Form: }
7x-5y=-16$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the equation of the line that passes through the given points $(
-3,-1
)$ and $(
2,6
),$ use the Two-Point Form of linear equations. Give the equation in the Slope-Intercept Form and in the Standard Form.
$\bf{\text{Solution Details:}}$
Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-Point Form of linear equations, where
\begin{array}{l}\require{cancel}
x_1=-3
,\\x_2=2
,\\y_1=-1
,\\y_2=6
,\end{array}
the equation of the line is
\begin{array}{l}\require{cancel}
y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)
\\\\
y-(-1)=\dfrac{-1-6}{-3-2}(x-(-3))
\\\\
y+1=\dfrac{-7}{-5}(x+3)
\\\\
y+1=\dfrac{7}{5}(x+3)
.\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+1=\dfrac{7}{5}(x+3)
\\\\
y+1=\dfrac{7}{5}(x)+\dfrac{7}{5}(3)
\\\\
y+1=\dfrac{7}{5}x+\dfrac{21}{5}
\\\\
y=\dfrac{7}{5}x+\dfrac{21}{5}-1
\\\\
y=\dfrac{7}{5}x+\dfrac{21}{5}-\dfrac{5}{5}
\\\\
y=\dfrac{7}{5}x+\dfrac{16}{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{7}{5}x+\dfrac{16}{5}
\\\\
5(y)=5\left( \dfrac{7}{5}x+\dfrac{16}{5} \right)
\\\\
5y=7x+16
\\\\
-7x+5y=16
\\\\
-1(-7x+5y)=-1(16)
\\\\
7x-5y=-16
.\end{array}
The equation of the line is
\begin{array}{l}\require{cancel}
\text{a) Slope-Intercept Form: }
y=\dfrac{7}{5}x+\dfrac{16}{5}
\\\\\text{b) Standard Form: }
7x-5y=-16
.\end{array}