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Math Help - Triangle ABC...

  1. #1
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    Triangle ABC...

    Triangle ABC is isosceles with BC=AC. The coordinates of the vertices are A(6, 1) and B(2, 8).
    Find the equation of the perpendicular bisector of AB.
    If the x coordinate of C is 3.5, find the y coordinate of C.
    Find the length of AB.
    Find the area of triangle ABC.
    Thanks in advance.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    a)
    1. Find the gradient of AB.
    2. Find the gradient perpendicular to that gradient.
    3. Find the midpoint of AB
    4. Use the point from 3. and the gradient fom 2. to get the perpendicular bisector of AB.

    b)
    1. Since this is an isosceles triangle, AC = BC. Use the distance formula:

    AC = BC

    AC^2 = (x_a - x_c)^2 + (y_a - y_c)^2

    BC^2 = (x_b - x_c)^2 + (y_b - y_c)^2

    (x_a, y_a) = (6, 1)\ , \ (x_b, y_b) = (2, 8)\ and \ (x_c, y_c) = (3.5, y_c)

    c)
    Use the distance formula again.

    AB^2 = (x_a - x_b)^2 + (y_a - y_b)^2

    d)
    1. You have the midpoint AB, label it X. Find the distance XC.

    2. Now, use Area\ of\ triangle = \dfrac12 \times base \times height

    Where base = AB, height = XC
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  3. #3
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    Thanks so much. That mostly makes perfect sense.
    But I don't completely understand what I'm meant to do for part b.
    Do you think you could try to help me through it??
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  4. #4
    MHF Contributor Unknown008's Avatar
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    As I told you AB = BC, hence;

    AB^2 = BC^2

    Replace by what I told you, and you get:

    (x_a - x_c)^2 + (y_a - y_c)^2 =     (x_b - x_c)^2 + (y_b - y_c)^2

    Substitute the values you know, to find the value of yc
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  5. #5
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    Quote Originally Posted by mandarep View Post
    Thanks so much. That mostly makes perfect sense.
    But I don't completely understand what I'm meant to do for part b.
    Do you think you could try to help me through it??
    TRY and show what you are able to do yourself.

    For this particular problem you don't really need to find those slopes. For an isosceles triangle, the perpendicular bisector of the base passes through the vertex angle. The perpendicular bisector of AB is the line through the midpoint of AB and C.
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  6. #6
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    Hello mandarp,
    I believe the best way to solve problems like these is to plot and do as much as possible without writing equations.In this particular case you would find out that the x value for C could be in error because it is very close to the midpoint of AB


    bjh
    Last edited by bjhopper; January 26th 2011 at 10:20 AM. Reason: mistake
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  7. #7
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    Hi guys.
    Thanks so much for trying to help me out.
    I ended up working through it, by substituting the values in.
    Sorry I just couldn't realise what to do before. Thanks again for everyone's help.
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