Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 70

Determine an equation of the line that satisfies the condition "through $(-1,3)$; parallel to $-x + 3y = 12$".

(a) Write the equation in slope intercept form.

We find the slope of the line $-x + 3y = 12$ and write it in slope intercept form


$
\begin{equation}
\begin{aligned}

-x + 3y =& 12
&& \text{Given equation}
\\
\\
3y =& x + 12
&& \text{Add each side by $x$}
\\
\\
y =& \frac{1}{3}x + \frac{12}{3}
&& \text{Divide each side by $3$}
\\
\\
y =& \frac{1}{3}x + 4
&&

\end{aligned}
\end{equation}
$



The slope is $\displaystyle \frac{1}{3}$. Using the Point Slope Form, with point $(-1,3)$, we have


$
\begin{equation}
\begin{aligned}


y - y_1 =& m(x - x_1)
&& \text{Point Slope Form}
\\
\\
y - 3 =& \frac{1}{3} [x - (-1)]
&& \text{Substitute } x = -1, y = 3 \text{ and } m = \frac{1}{3}
\\
\\
y - 3 =& \frac{1}{3}x + \frac{1}{3}
&& \text{Distributive Property}
\\
\\
y =& \frac{1}{3}x + \frac{10}{3}
&& \text{Add each side by $3$}
\\
\\
y =& \frac{1}{3}x + \frac{10}{3}
&& \text{Slope Intercept Form}


\end{aligned}
\end{equation}
$


(b) Write the equation in standard form.


$
\begin{equation}
\begin{aligned}

y =& \frac{1}{3}x + \frac{10}{3}
&& \text{Slope Intercept Form}
\\
\\
3y =& x + 10
&& \text{Multiply each side by $3$}
\\
\\
-x + 3y =& 10
&& \text{Standard Form}

\end{aligned}
\end{equation}
$