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Math Help - Perpendicular Bisector

  1. #1
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    Perpendicular Bisector

    Hey guys, something that is probably easy for you is lurking below:

    We're working on Perpendicular bisectors at the minute, and all my working is perfectly right, but I still seem to be getting the wrong answer. I need to get a hang of it by tomorrow morning (It's currently 10pm) as we are moving on and won't be covering it again.

    The exercises which we are using are:

    1) Find the equation of the perpendicular bisector of the line joining each pair of points:

    A(5, -7) and J (6,4)


    I know that I need to get to y-b = m(x-a) and to do this I must first need to figure out the middle point using:

    x1+x2 / 2

    y1+y2
    / 2

    Which will fill in a and b of y-b = m(x-a)

    I then need to use:

    m = y2-y1 / x2-x1


    then using the answer to this, I then need to use m1 x m2 = -1 to be able to fill in M in the y-b = m(x-a).

    I then need to switch it around so that x + y + n = 0
    (where n is number)

    But I still seem to be getting the wrong answer. If someone could show me the solution to the above one, I might be able to figure out where I am going wrong.

    Also, I have no idea where to start with:

    Find the equation of the perpendicular bisector of each side of a triangle with vertices:

    P (1, 3), R (0, 4) and Q (5, 2)


    Any help would be much appreciated. I think that once I get a grasp of where I am going wrong with the first part, I should be able to figure out the second question.

    Thank you so much in advance!
    Last edited by Mathy; September 5th 2007 at 01:10 PM. Reason: Layout
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathy View Post
    1) Find the equation of the perpendicular bisector of the line joining each pair of points:

    A(5, -7) and J (6,4)
    Find the midpoint:
    x = \frac{5 + 6}{2} = \frac{11}{2}

    y = \frac{-7 + 4}{2} = -\frac{3}{2}

    Now find the slope of the line connecting points A and J:
    m = \frac{-7 - 4}{5 - 6} = 11

    So the line we want has a slope of -\frac{1}{11} and intersects the point \left ( \frac{11}{2}, -\frac{3}{2} \right ) .

    So
    y = mx + b

    -\frac{3}{2} = -\frac{1}{11} \cdot \frac{11}{2} + b

    So
    b = -\frac{3}{2} + \frac{1}{11} \cdot \frac{11}{2}

    b = -\frac{3}{2} + \frac{1}{2} = -1

    So your perpendicular bisector is
    y = -\frac{1}{11}x - 1

    The graph below shows everything. (Sorry the bisector doesn't look perpendicular. The axes are scaled differently and I couldn't get it to "square up" nicely.)

    -Dan
    Attached Thumbnails Attached Thumbnails Perpendicular Bisector-bisector.jpg  
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