Question

14. PQ is formed by P(10, 4) and Q(2,-8). If line k is the perpendicular bisector of PQ, write a linear equation for k in slope-intercept form.

Answer 1

First, let's find the equation for PQ:

$\begin{gathered} \text{Let:} \\ (x1,y1)=(10,4) \\ (x2,y2)=(2,-8) \\ m=\frac{y2-y1}{x2-x1}=\frac{-8-4}{2-10}=\frac{-12}{-8}=\frac{3}{2} \end{gathered}$

Using the point-slope equation:

$\begin{gathered} y-y1=m(x-x1) \\ y-4=\frac{3}{2}(x-10) \\ y-4=\frac{3}{2}x-15 \\ y=\frac{3}{2}x-11 \end{gathered}$

If line k is the perpendicular bisector of PQ, we need to find the middle point, so:

$\begin{gathered} xm=\frac{x1+x2}{2} \\ xm=\frac{10+2}{2}=\frac{12}{2}=6 \\ ym=\frac{y1+y2}{2} \\ ym=\frac{4-8}{2}=-\frac{4}{2}=-2 \end{gathered}$

Since k is perpendicular to PQ:

$\begin{gathered} m1\times m2=-1 \\ \text{where:} \\ m1=\frac{3}{2} \\ m2=-\frac{1}{m1} \\ m2=-\frac{2}{3} \end{gathered}$

Using the middle point and the slope m2:

$\begin{gathered} y-ym=m2(x-xm) \\ y-(-2)=-\frac{2}{3}(x-6) \\ y+2=-\frac{2}{3}x+4 \\ y=-\frac{2}{3}x+2 \end{gathered}$