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=(y2-y1)/(x2-x1)=(-8-4)/(2-10)=(-12)/(-8)=(3)/(2) \end{gathered}`

Using the point-slope equation:

`\begin{gathered} y-y1=m(x-x1) \\ y-4=(3)/(2)(x-10) \\ y-4=(3)/(2)x-15 \\ y=(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=(x1+x2)/(2) \\ xm=(10+2)/(2)=(12)/(2)=6 \\ ym=(y1+y2)/(2) \\ ym=(4-8)/(2)=-(4)/(2)=-2 \end{gathered}`

Since k is perpendicular to PQ:

`\begin{gathered} m1* m2=-1 \\ \text{where:} \\ m1=(3)/(2) \\ m2=-(1)/(m1) \\ m2=-(2)/(3) \end{gathered}`

Using the middle point and the slope m2:

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