Precalculus, Chapter 1, 1.3, Section 1.3, Problem 62

Determine the equation of the line that is parallel to the line $x - 2y = -5$ containing the point $(0,0)$. Express your answer using the general form or the slope intercept form of the equation of a line, which ever you prefer.

Since the two lines are parallel, the slope of the line that we
need to find equals the slope of the line $x - 2y = -5$. We start by writing the equation $x - 2y = -5$ in slope-intercept form.


$
\begin{equation}
\begin{aligned}

x-2y =& -5
\\
-2y =& -x-5
\\
y =& \frac{1}{2}x + \frac{5}{2}

\end{aligned}
\end{equation}
$


The slope is $\displaystyle \frac{1}{2}$. The other equation should have a slope $\displaystyle \frac{1}{2}$ and contains the point $(0,0)$. By using Point Slope Form to find the equation


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x- x_1)
&& \text{Point Slope Form}
\\
\\
y - 0 =& \frac{1}{2} (x-0)
&& \text{Substitute } m = \frac{1}{2}, x = 0 \text{ and } y = 0
\\
\\
y =& \frac{1}{2}x
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$