Question

Find an equation for the perpendicular bisector of the line segment whose endpoints are (6, -3) and (2,5).

Answer 1

The line segment has points (6,-3) and (2,5).

With these ponts we must find the middle point. Generally, this is given by

$\text{midpoint}=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

where

$\begin{gathered} (x_1,y_1)=\mleft(6,-3\mright) \\ (x_2,y_2)=(2,5) \end{gathered}$

By substituying these point into the formula, we have

$\begin{gathered} \text{midpoint}=(\frac{6+2}{2},\frac{-3+5}{2}) \\ \text{midpoint}=(\frac{8}{2},\frac{2}{2}) \\ \text{midpoint}=(4,1) \end{gathered}$

The perpendicular bisector must pass through point (4,1) and must have a perpendicular slope to the given

line segment. This line segment has slope:

$\begin{gathered} m=\frac{5-(-3)}{2-6} \\ m=\frac{5+3}{-4} \\ m=-\frac{8}{4} \\ m=-2 \end{gathered}$

hence, the perpendicular bisector has the "negative reciprocal" value of m. That is,

$m_p=-\frac{1}{m}$

therefore, our perpendicular bisector has slope:

$\begin{gathered} m_p=-\frac{1}{-2} \\ \text{hence, } \\ m_p=\frac{1}{2} \end{gathered}$

Finally, the perpendicular bisector has the form:

$y=m_px+b$

we know the slope mp. The y-intercept b can be computed by means of the middle point (4,1).

This can be done as

$\begin{gathered} y=m_px+b\text{ implies} \\ 1=\frac{1}{2}(4)+b \\ 1=2+b \\ b=1-2 \\ b=-1 \end{gathered}$

Therefore, the perpendicular bisector is

$y=\frac{1}{2}x-1$