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}=((x_1+x_2)/(2),(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}=((6+2)/(2),(-3+5)/(2)) \\ \text{midpoint}=((8)/(2),(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=(5-(-3))/(2-6) \\ m=(5+3)/(-4) \\ m=-(8)/(4) \\ m=-2 \end{gathered}`

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

`m_p=-(1)/(m)`

therefore, our perpendicular bisector has slope:

`\begin{gathered} m_p=-(1)/(-2) \\ \text{hence, } \\ m_p=(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=(1)/(2)(4)+b \\ 1=2+b \\ b=1-2 \\ b=-1 \end{gathered}`

Therefore, the perpendicular bisector is

`y=(1)/(2)x-1`