Precalculus, Chapter 1, 1.3, Section 1.3, Problem 68

Determine the equation of the line that is perpendicular to the line $x - 2y = -5$ containing the point $(0,4)$. Express your answer using the general form or the slope intercept form of the equation of a line, which ever you prefer.

We first write the given equation in slope intercept form to find its slope


$
\begin{equation}
\begin{aligned}

x - 2y =& -5
&&
\\
-2y =& -x-5
&& \text{Solve for } y
\\
y =& \frac{1}{2}x + \frac{5}{2}
&& \text{Write in the form } y = mx+b

\end{aligned}
\end{equation}
$


The slope is $\displaystyle \frac{1}{2}$. Any line perpendicular to this line will have slope $-2$. The point $(0,4)$ is on the line with slope $-2$, we use the point slope form to find this equation.


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x- x_1)
&& \text{Point Slope Form}
\\
y - 4 =& -2(x-0)
&& \text{Substitute } m = -2, x = 0 \text{ and } y = 4
\\
y =& -2x+4
&& \text{Slope Intercept Form}
\\
\text{or} &
&&
\\
2x+y =& 4
&& \text{General Form}

\end{aligned}
\end{equation}
$