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Thread: 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 . . .


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    . . . There!

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  3. #3
    Junior Member
    Joined
    Oct 2005
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    Smile

    Thanks a lot!
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