1. ## perpenicular bisector...?

find the equation of the perpendicular bisector of (2,-5) and (6,9), expressing your answer in the form: ax + by + c = 0

thanks for any help i can quite seem to get this one...

and its in for tomorrow!! gah!

2. Originally Posted by pop_91
find the equation of the perpendicular bisector of (2,-5) and (6,9), expressing your answer in the form: ax + by + c = 0

thanks for any help i can quite seem to get this one...

and its in for tomorrow!! gah!
Find the midpoint of the line segment through (2, -5) and (6, 9):

$Midpoint=\left(\frac{2+6}{2} \ \ , \ \ \frac{9-5}{2}\right)=(4, 2)$

Find the slope of the line through (2, -5) and (6, 9).

slope = $\frac{9-^-5}{6-2}=\frac{14}{4}=\frac{7}{2}$

The slope of the perpendicular bisector is the negative reciprocal of that slope, so it would be $-\frac{2}{7}$ and pass through the midpoint (4, 2).

Using y=mx+b and m = $-\frac{2}{7}$ through (4, 2).

$2=-\frac{2}{7}(4)+b$

$\frac{22}{7}=b$

$y=-\frac{2}{7}x+\frac{22}{7}$

$7y=-2x+22$

$2x+7y-22=0$

3. thank you soo much!!!!

4. Hello, pop_91!

All the necessary information is there ... just dig it up.

Find the equation of the perpendicular bisector of $A(2,-5)$ and $B(6,9),$
expressing your answer in the form: $ax + by + c \:=\: 0$

The perpendicular bisector of $AB$ goes through the midpoint of $AB$
. . and is perpendicular to $AB.$

The midpoint of $AB$ is: . $\left(\frac{2+6}{2},\;\frac{\text{-}5+9}{2}\right) \;=\;(4,\:2)$

The slope of $AB$ is: . $m_1 \;=\;\frac{9-(\text{-}5)}{6-2} \:=\:\frac{14}{4} \:=\:\frac{7}{2}$
The perpendicular slope is: . $m_2 \:=\:\text{-}\frac{2}{7}$

The line through $(4,\:2)$ with slope $\text{-}\frac{2}{7}$ is:
. . $y - 2 \;=\;\text{-}\frac{2}{7}(x - 4) \quad\Rightarrow\quad y - 2 \;=\;\text{-}\frac{2}{7}x + \frac{8}{7}$

Multiply by 7: . $7y - 14 \;=\;\text{-}2x + 8$

. . Therefore: . ${\color{blue}2x + 7y - 22 \;=\;0}$