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}
$