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}
$