Results 1 to 7 of 7
Like Tree3Thanks
  • 2 Post By Plato
  • 1 Post By Prove It

Math Help - Perpendicular bisectors

  1. #1
    Junior Member
    Joined
    Jan 2013
    From
    London
    Posts
    30

    Perpendicular bisectors

    I can't seem to find a formula to find the perpendicular bisector of a line. Anyone know what it is?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,607
    Thanks
    1574
    Awards
    1

    Re: Perpendicular bisectors

    Quote Originally Posted by alexpasty2013 View Post
    I can't seem to find a formula to find the perpendicular bisector of a line. Anyone know what it is?

    Suppose each of P: (x_1,y_1)~\&~Q: (x_2,y_2) is a point and y_1\ne y_2.

    The perpendicular bisector of \overline{PQ} is y = \left( { - \frac{{{x_1} - {x_2}}}{{{y_1} - {y_2}}}} \right)\left( {x - \frac{{{x_1} + {x_2}}}{2}} \right) + \left( {\frac{{{y_1} + {y_2}}}{2}} \right)
    Thanks from MarkFL and alexpasty2013
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2013
    From
    London
    Posts
    30

    Re: Perpendicular bisectors

    So we have 2 points A (-2, 2) and B (3 , 1) the the perpendicular bisector is;

    AB is y = (-(-2)-3/-2+1) (x- (-2)+3/2) + (-2+1/2) =

    y = -5/3(x-1/2)+ (-1/2)

    and our final equation is

    y= -5x-1

    is that right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,487
    Thanks
    1391

    Re: Perpendicular bisectors

    Quote Originally Posted by alexpasty2013 View Post
    So we have 2 points A (-2, 2) and B (3 , 1) the the perpendicular bisector is;

    AB is y = (-(-2)-3/-2+1) (x- (-2)+3/2) + (-2+1/2) =

    y = -5/3(x-1/2)+ (-1/2)

    and our final equation is

    y= -5x-1

    is that right?
    The slope of the line through A and B is \displaystyle \begin{align*} \frac{1 - 2}{3 - (-2)} = -\frac{1}{5} \end{align*}. So the slope of your perpendicular bisector is 5, since the two slopes will multiply to give -1.

    The midpoint of AB is \displaystyle \begin{align*} \left( \frac{-2 + 3}{2} , \frac{ 2 + 1}{2} \right) = \left( \frac{ 1}{2} , \frac{3}{2} \right) \end{align*}

    So the perpendicular bisector is of the form \displaystyle \begin{align*} y = mx + c \end{align*} with \displaystyle \begin{align*} m = 5 \end{align*} and a point it passes through is \displaystyle \begin{align*} (x, y) = \left( \frac{1}{2} , \frac{3}{2} \right) \end{align*}. Substituting these gives

    \displaystyle \begin{align*} \frac{3}{2} &= 5 \left( \frac{1}{2} \right) + c \\ \frac{3}{2} &= \frac{5}{2} + c \\ c &= -1 \end{align*}

    and so the equation of your perpendicular bisector is \displaystyle \begin{align*} y = 5x - 1 \end{align*}.
    Thanks from alexpasty2013
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2013
    From
    London
    Posts
    30

    Re: Perpendicular bisectors

    So I got the right answer by the wrong means?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2013
    From
    London
    Posts
    30

    Re: Perpendicular bisectors

    this is wrong - the point (1/2, 3, 2) does not lie on the line, my original point (0.5, -0.5) is on the line and apparently in the centre
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jul 2012
    From
    INDIA
    Posts
    828
    Thanks
    209

    Re: Perpendicular bisectors

    You cannot find the bisector of a line for you cannot measure a line.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 30th 2012, 12:12 PM
  2. Replies: 2
    Last Post: April 17th 2010, 03:04 PM
  3. About Perpendicular Bisectors
    Posted in the Geometry Forum
    Replies: 2
    Last Post: December 25th 2009, 11:19 AM
  4. Perpendicular bisectors and centres of circles
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: March 6th 2009, 06:49 AM
  5. perpendicular bisectors and circles
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 16th 2006, 01:29 PM

Search Tags


/mathhelpforum @mathhelpforum