Results 1 to 4 of 4

Thread: perpenicular bisector...?

  1. #1
    Junior Member
    Joined
    Aug 2008
    From
    England
    Posts
    40

    Unhappy 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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,550
    Thanks
    15
    Awards
    1
    Quote Originally Posted by pop_91 View Post
    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).

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

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

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

    $\displaystyle 7y=-2x+22$

    $\displaystyle 2x+7y-22=0$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2008
    From
    England
    Posts
    40
    thank you soo much!!!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    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: . $\displaystyle {\color{blue}2x + 7y - 22 \;=\;0}$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Perpendicular bisector
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Oct 11th 2010, 04:29 PM
  2. Perpendicular Bisector?
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Jul 10th 2010, 05:03 AM
  3. Bisector
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Mar 15th 2010, 04:53 AM
  4. Bisector
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Feb 19th 2010, 11:38 PM
  5. perpenicular equation of line
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Apr 9th 2007, 03:43 PM

Search Tags


/mathhelpforum @mathhelpforum