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}$