Answer
$\text{a) Slope-Intercept Form: }
y=-\dfrac{4}{3}x+\dfrac{29}{3}
\\\\\text{b) Standard Form: }
4x+3y=29$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the equation of the line with the following properties:
\begin{array}{l}\require{cancel}
\text{Slope: }
-\dfrac{4}{3}
\\\text{Through: }
(2,7)
,\end{array}
use the Point-Slope Form of linear equations. Give the equation in the Slope-Intercept Form and in the Standard Form.
$\bf{\text{Solution Details:}}$
Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions,
\begin{array}{l}\require{cancel}
y_1=7
,\\x_1=2
,\\m=-\dfrac{4}{3}
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-7=-\dfrac{4}{3}(x-2)
.\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-7=-\dfrac{4}{3}(x-2)
\\\\
y-7=-\dfrac{4}{3}(x)-\dfrac{4}{3}(-2)
\\\\
y-7=-\dfrac{4}{3}x+\dfrac{8}{3}
\\\\
y=-\dfrac{4}{3}x+\dfrac{8}{3}+7
\\\\
y=-\dfrac{4}{3}x+\dfrac{8}{3}+\dfrac{21}{3}
\\\\
y=-\dfrac{4}{3}x+\dfrac{29}{3}
.\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{4}{3}x+\dfrac{29}{3}
\\\\
3(y)=3\left( -\dfrac{4}{3}x+\dfrac{29}{3} \right)
\\\\
3y=-4x+29
\\\\
4x+3y=29
.\end{array}
The equation of the line is
\begin{array}{l}\require{cancel}
\text{a) Slope-Intercept Form: }
y=-\dfrac{4}{3}x+\dfrac{29}{3}
\\\\\text{b) Standard Form: }
4x+3y=29
.\end{array}
