Intermediate Algebra, Chapter 3, Summary Exercises, Section Summary Exercises, Problem 16

Write an equation of the line "through $(-3,6)$ with slope $\displaystyle \frac{2}{3}$".

(a) In slope-intercept form

Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
&& \text{Point Slope Form}
\\
\\
y - 6 =& \frac{2}{3} [x - (-3)]
&& \text{Substitute } x = -3, y = 6 \text{ and } m = \frac{2}{3}
\\
\\
y - 6 =& \frac{2}{3}x + 2
&& \text{Distributive Property}
\\
\\
y =& \frac{2}{3}x + 2 + 6
&& \text{Add each side by $6$}
\\
\\
y =& \frac{2}{3}x + 8
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$



(b) In standard form


$
\begin{equation}
\begin{aligned}

y =& \frac{2}{3}x + 8
&& \text{Slope Intercept Form}
\\
\\
- \frac{2}{3}x + y =& 8
&& \text{Standard Form}
\\
\\
\text{or} &
&&
\\
\\
-2x+ 3y =& 24
&&

\end{aligned}
\end{equation}
$