Precalculus, Chapter 1, 1.3, Section 1.3, Problem 66

Determine the equation of the line that is perpendicular to the line $y = 2x-3$ containing the point $(1,-2)$. 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 perpendicular, the product of their slopes must be $-1$. We use the given equation to find the slope. The line $y = 2x-3$ has a slope $2$. So, any line perpendicular to this line will have slope $\displaystyle \frac{-1}{2}$. By using the point slope form to find the equation with point $(1,-2)$. We have,


$\begin{equation} \begin{aligned} y - y_1 =& m (x- x_1) && \text{Point Slope Form} \\ \\ y - (-2) =& \frac{-1}{2} (x-1) && \text{Substitute } m = \frac{-1}{2}, x = 1 \text{ and } y = -2 \\ \\ y =& \frac{-1}{2}x + \frac{1}{2} - 2 && \text{Simplify} \\ y =& \frac{-1}{2}x - \frac{3}{2} && \text{Slope Intercept Form} \\ \\ \text{or} & && \\ \\ \frac{1}{2}x + y =& \frac{-3}{2} && \text{General Form} \end{aligned} \end{equation}$