Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 40

Determine an equation of the line that satisfies the condition "through $(7,-2)$; slope $\displaystyle \frac{1}{4}$".

(a) Write the equation in standard form.

Use the Point Slope Form of the equation of a line with $(x_1,y_1) = (7,-2)$ and $m = \displaystyle \frac{1}{4}$


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
&& \text{Point Slope Form}
\\
\\
y - (-2) =& \frac{1}{4} (x-7)
&& \text{Substitute $x = 7, y = -2$ and } m = \frac{1}{4}
\\
\\
y + 2 =& \frac{1}{4}x - \frac{7}{4}
&& \text{Distributive Property}
\\
\\
- \frac{1}{4}x + y =& - \frac{7}{4} - 2
&& \text{Subtract each side by } \left( \frac{1}{4}x + 2 \right)
\\
\\
- \frac{1}{4}x + y =& - \frac{15}{4}
&& \text{Standard Form}
\\
\text{or} &
&&&
\\
x - 4y =& 15
&&

\end{aligned}
\end{equation}
$



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


$
\begin{equation}
\begin{aligned}

- \frac{1}{4}x + y =& - \frac{15}{4}
&& \text{Standard Form}
\\
\\
y =& \frac{1}{4}x - \frac{15}{4}
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$