Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 44

Determine an equation of the line that satisfies the condition "through $(6,-1.2)$; slope $0.8$".

(a) Write the equation in standard form.

Use the Point Slope Form of the equation of a line with $(x_1,y_1) = (6,-1.2)$ and $m = 0.8$


$\begin{equation} \begin{aligned} y - y_1 =& m (x - x_1) && \text{Point Slope Form} \\ y - (-1.2) =& 0.8(x - 6) && \text{Substitute $x = 6, y = -1.2$ and } m = 0.8 \\ y + 1.2 =& 0.8x - 4.8 && \text{Distributive Property} \\ -0.8x + y =& -4.8 - 1.2 && \text{Subtract each side by $(0.8x + 1.2)$} \\ -0.8x + y =& -6 && \text{Standard Form} \\ \text{or} & && \\ 0.8x - y =& 6 && \end{aligned} \end{equation}$



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


$\begin{equation} \begin{aligned} -0.8x + y =& -6 && \text{Standard Form} \\ y =& 0.8x - 6 && \text{Slope Intercept Form} \end{aligned} \end{equation}$