Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 38

Determine an equation of the line that satisfies the condition "through $(-1,6)$; slope $\displaystyle - \frac{5}{6}$".

(a) Write the equation in standard form.

Use the Point Slope Form of the equation of a line with $(x_1,y_1) = (-1,6)$ and $m = \displaystyle - \frac{5}{6}$


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
&& \text{Point Slope Form}
\\
\\
y - 6 =& - \frac{5}{6} [x - (-1)]
&& \text{Substitute $x = -1, y = 6$ and } m = - \frac{5}{6}
\\
\\
y - 6 =& - \frac{5}{6}x - \frac{5}{6}
&& \text{Distributive Property}
\\
\\
6y - 36 =& -5x - 5
&& \text{Multiply each side by $6$}
\\
\\
5x + 6y =& -5 + 36
&& \text{Add each side by $(5x + 36)$}
\\
\\
5x + 6y =& 31
&& \text{Standard Form}

\end{aligned}
\end{equation}
$



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


$
\begin{equation}
\begin{aligned}

5x + 6y =& 31
&& \text{Standard Form}
\\
\\
6y =& -5x + 31
&& \text{Subtract each side by $5x$}
\\
\\
y =& - \frac{5}{6}x + \frac{31}{6}
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$