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):
$\displaystyle Midpoint=\left(\frac{2+6}{2} \ \ , \ \ \frac{9-5}{2}\right)=(4, 2)$
Find the slope of the line through (2, -5) and (6, 9).
slope = $\displaystyle \frac{9-^-5}{6-2}=\frac{14}{4}=\frac{7}{2}$
The slope of the perpendicular bisector is the negative reciprocal of that slope, so it would be $\displaystyle -\frac{2}{7}$ and pass through the midpoint (4, 2).
Using y=mx+b and m = $\displaystyle -\frac{2}{7}$ through (4, 2).
$\displaystyle 2=-\frac{2}{7}(4)+b$
$\displaystyle \frac{22}{7}=b$
$\displaystyle y=-\frac{2}{7}x+\frac{22}{7}$
$\displaystyle 7y=-2x+22$
$\displaystyle 2x+7y-22=0$
Hello, pop_91!
All the necessary information is there ... just dig it up.
Find the equation of the perpendicular bisector of $\displaystyle A(2,-5)$ and $\displaystyle B(6,9),$
expressing your answer in the form: $\displaystyle ax + by + c \:=\: 0$
The perpendicular bisector of $\displaystyle AB$ goes through the midpoint of $\displaystyle AB$
. . and is perpendicular to $\displaystyle AB.$
The midpoint of $\displaystyle AB$ is: .$\displaystyle \left(\frac{2+6}{2},\;\frac{\text{-}5+9}{2}\right) \;=\;(4,\:2)$
The slope of $\displaystyle AB$ is: .$\displaystyle m_1 \;=\;\frac{9-(\text{-}5)}{6-2} \:=\:\frac{14}{4} \:=\:\frac{7}{2}$
The perpendicular slope is: .$\displaystyle m_2 \:=\:\text{-}\frac{2}{7}$
The line through $\displaystyle (4,\:2)$ with slope $\displaystyle \text{-}\frac{2}{7}$ is:
. . $\displaystyle y - 2 \;=\;\text{-}\frac{2}{7}(x - 4) \quad\Rightarrow\quad y - 2 \;=\;\text{-}\frac{2}{7}x + \frac{8}{7}$
Multiply by 7: .$\displaystyle 7y - 14 \;=\;\text{-}2x + 8$
. . Therefore: . $\displaystyle {\color{blue}2x + 7y - 22 \;=\;0}$