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Math Help - balance point in a triangle

  1. #1
    Junior Member
    Joined
    Oct 2005
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    Smile balance point in a triangle

    Hey guys,
    I have a work sheet to do and I did almost all the questions except 9.
    http://i13.tinypic.com/4c7nz8h.jpg
    I need you help how to do that.
    I added all the answers from the previous questions in this sheet.

    A(0,6) B(0,-2) C(12,2)
    mid point A to B= (0,2)
    mid point A to C= (6,4)
    mid point B to C= (6,0)
    Intersection point(4,2)

    Thanks a lot!
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, dgolverk!

    Your intersection point is incorrect . . .


    A(0,6),\;B(0,-2),\; C(12,2)

    \begin{array}{ccc}\text{midpoint of }AB = D(0,2) \\ \text{midpoint of }BC = E(6,0) \\ \text{midpoint of }AC = F(6,4)\end{array}

    \text{Intersection point: }P(4,2) . . . no

    Since AB is vertical and its midpoint is D(0,2),
    . . its perpendicular bisector is the horizontal line: y \,= \,2

    The slope of AC is: . m \:=\:\frac{2 - 6}{12 - 0} \:=\:\frac{-4}{12}\:=\:-\frac{1}{3}

    The perpendicular slope is: . M_{\perp} = +3

    The line through F(6,4) with slope 3 is:
    . . y - 4\:=\:3(x - 6)\quad\Rightarrow\quad y \:=\:3x-14

    The two perpendicular bisectors: y \:=\:3x - 14 and y \,= \,2
    . . intersect when: . 3x - 14 \:=\:2\quad\Rightarrow\quad x = \frac{16}{3}

    Therefore, the intersection point is: . \boxed{P\left(\frac{16}{3},\,2\right)}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We must verify that P is equidistant from A,\,B,\,C.

    Distance formula
    The distance between (x_1,\,y_1) and (x_2,\,y_2) is: . d \;=\;\sqrt{x_2-x_1)^2+(y_2-y_1)^2}

    Distance from P\left(\frac{16}{3},\,2\right) to A(0,6)
    . . PA \:=\:\sqrt{\left(\frac{16}{3} - 0\right)^2 + (2 - 6)^2} \:=\:\sqrt{\frac{256}{9} + 16} \:=\:\sqrt{\frac{400}{9}} \:=\:\boxed{\frac{20}{3}}

    Distance from P\left(\frac{16}{3},\,2\right) to B(-,-2)
    . . PB \:=\:\sqrt{\left(\frac{16}{3} - 0\right)^2 + \left(2 - [-2]\right)^2} \:=\:\sqrt{\frac{256}{9} + 16} \:=\:\sqrt{\frac{400}{9}} \:=\:\boxed{\frac{20}{3}}

    Distance from P\left(\frac{16}{3},\,2\right) to C(12,2)
    . . PC \:=\:\sqrt{\left(\frac{16}{3} - 12\right)^2 + (2 - 2)^2} \;=\;\sqrt{\left(\frac{-20}{3}\right)^2 + 0^2} \:=\:\sqrt{\frac{400}{9}} \:=\:\boxed{\frac{20}{3}}

    . . . There!

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  3. #3
    Junior Member
    Joined
    Oct 2005
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    Smile

    Thanks a lot!
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