Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 72

Determine an equation of the line that satisfies the condition "through $(2,-7)$; perpendicular to $5x + 2y = 18$".

(a) Write the equation in slope intercept form.

We find the slope of the line $5x + 2y = 18$ and write it in slope intercept form


$
\begin{equation}
\begin{aligned}

5x + 2y =& 18
&& \text{Given equation}
\\
\\
2y =& -5x + 18
&& \text{Subtract each side by $5x$}
\\
\\
y =& - \frac{5}{2}x + 9
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$


To be perpendicular to the line $\displaystyle y = - \frac{5}{2}x + 9$, a line must have a slope that is the negative reciprocal of $\displaystyle - \frac{5}{2}$ and that is $\displaystyle \frac{2}{5}$. Using Point Slope Form


$
\begin{equation}
\begin{aligned}


y - y_1 =& m(x - x_1)
&& \text{Point Slope Form}
\\
\\
y - (-7) =& \frac{2}{5} (x-2)
&& \text{Substitute } x = 2, y = -7 \text{ and } m = \frac{2}{5}
\\
\\
y + 7 =& \frac{2}{5}x - \frac{4}{5}
&& \text{Distributive Property}
\\
\\
y =& \frac{2}{5}x - \frac{4}{5} - 7
&& \text{Subtract each side by $7$}
\\
\\
y =& \frac{2}{5}x - \frac{39}{5}
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$


(b) Write the equation in standard form.


$
\begin{equation}
\begin{aligned}

y =& \frac{2}{5}x - \frac{39}{5}
&& \text{Slope Intercept Form}
\\
\\
5y =& 2x - 39
&& \text{Multiply each side by $5$}
\\
\\
-2x + 2y =& -39
&& \text{Standard Form}
\\
\text{or} &
&&
\\
2x - 2y =& 39
&&

\end{aligned}
\end{equation}
$