Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 66

Determine an equation of the line passing through the points $\displaystyle \left( \frac{3}{4}, \frac{8}{3} \right)$ and $\displaystyle \left( \frac{2}{5}, \frac{2}{3} \right)$.

(a) Write the equation in standard form.

Using the Slope Formula,

$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\displaystyle \frac{2}{3} - \frac{8}{3}}{\displaystyle \frac{2}{5} - \frac{3}{4}} = \frac{\displaystyle - \frac{6}{3}}{\displaystyle - \frac{7}{20}} = - \frac{6}{3} \cdot - \frac{20}{7} = \frac{40}{7}$

Using Point Slope Form, where $m = \displaystyle \frac{40}{7}$ and $(x_1,y_1) = \displaystyle \left( \frac{3}{4}, \frac{8}{3} \right)$



$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
&& \text{Point Slope Form}
\\
\\
y - \frac{8}{3} =& \frac{40}{7} \left( x - \frac{3}{4} \right)
&& \text{Substitute } x = \frac{3}{4}, y = \frac{8}{3} \text{ and } m = \frac{40}{7}
\\
\\
y - \frac{8}{3} =& \frac{40}{7}x - \frac{30}{7}
&& \text{Distributive Property}
\\
\\
21y - 56 =& 120x - 90
&& \text{Multiply each side by $21$}
\\
\\
-120x + 21y =& -90 + 56
&& \text{Subtract each side by $(120x - 56)$}
\\
\\
-120x + 21y =& -34
&& \text{Standard Form}
\\
\text{or} &
&&
\\
120x - 21y =& 34
&&

\end{aligned}
\end{equation}
$



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


$
\begin{equation}
\begin{aligned}

-120x + 21y =& -34
&& \text{Standard Form}
\\
\\
21y =& 120x - 34
&& \text{Add each side by $120x$}
\\
\\
y =& \frac{120}{21}x - \frac{34}{21}
&& \text{Divide each side by $21$}
\\
\\
y =& \frac{40}{7}x - \frac{34}{21}
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$