Question

what is the equation of the line perpendicular to $overleftrightarrow{ab}$ that passes through point $p$?
a. $y = \frac{3}{2}x+\frac{1}{2}$
b. $y = -\frac{2}{3}x-\frac{2}{3}$
c. $y = \frac{2}{3}x+\frac{14}{3}$
d. $y = -\frac{3}{2}x - 4$

Explanation:

Step1: Find slope of line AB

Let \(A(-4, 2)\) and \(B(2,-2)\). Slope \(m_{AB}=\frac{y_B - y_A}{x_B - x_A}=\frac{-2 - 2}{2+4}=\frac{-4}{6}=-\frac{2}{3}\).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is - 1. Let the slope of the line perpendicular to \(AB\) be \(m\). Then \(m\times(-\frac{2}{3})=-1\), so \(m = \frac{3}{2}\).

Step3: Use point - slope form

Assume point \(P(1,2)\). The point - slope form of a line is \(y - y_1=m(x - x_1)\). Substituting \(m=\frac{3}{2}\), \(x_1 = 1\) and \(y_1=2\) gives \(y - 2=\frac{3}{2}(x - 1)\).

Step4: Convert to slope - intercept form

\[

$$\begin{align*} y-2&=\frac{3}{2}x-\frac{3}{2}\\ y&=\frac{3}{2}x-\frac{3}{2}+2\\ y&=\frac{3}{2}x+\frac{1}{2} \end{align*}$$

\]

Answer:

A. \(y=\frac{3}{2}x+\frac{1}{2}\)