Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 58

Determine an equation of the line passing through the points $(-2,5)$ and $(-8,1)$.

(a) Write the equation in standard form.

Using the Slope Formula,

$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 5}{-8-(-2)} = \frac{-4}{-6} = \frac{2}{3}$

Using Point Slope Form, where $m = \displaystyle \frac{2}{3}$ and $(x_1,y_1) = (-2,5)$



$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
&& \text{Point Slope Form}
\\
\\
y - 5 =& \frac{2}{3} [x - (-2)]
&& \text{Substitute } x = -2, y = 5 \text{ and } m = \frac{2}{3}
\\
\\
y - 5 =& \frac{2}{3}x + \frac{4}{3}
&& \text{Distributive Property}
\\
\\
3y - 15 =& 2x + 4
&& \text{Multiply each side by $3$}
\\
\\
-2x + 3y =& 4 + 15
&& \text{Subtract each side by $(2x - 15)$}
\\
\\
-2x + 3y =& 19
&& \text{Standard Form}

\end{aligned}
\end{equation}
$



(b) Write the equation in slope-intercept form.


$
\begin{equation}
\begin{aligned}

-2x + 3y =& 19
&& \text{Standard Form}
\\
\\
3y =& 2x + 19
&& \text{Add each side by $2x$}
\\
\\
y =& \frac{2}{3}x + \frac{19}{3}
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$