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.


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