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Math Help - Area of a triangle given the vertices.

  1. #1
    Member rtblue's Avatar
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    Area of a triangle given the vertices.

    What is the area of the triangle whose vertices are (0, 0), (14, 6), and (12, 9)?

    I realize that you could use Distance formula, followed by Hero's Formula to obtain the answer, but i'm sure there is a faster way to go about this (that way has too many radicals and fractions, and is easy to mess up on).

    Suggestions to get me started?
    Last edited by mr fantastic; March 25th 2010 at 06:49 PM. Reason: Changed post title
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  2. #2
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    Hello, rtblue!

    As a matter of fact, there is a neat formula for this problem.


    What is the area of the triangle whose vertices are (0, 0), (14, 6), and (12, 9)?
    If you understand determinants, you'll like this . . .


    The area of the triangle with vertices: . (x_1,y_1),\:(x_2,y_2),\:(x_3,y_3) is:

    . . . . . A \;=\;\frac{1}{2}\begin{vmatrix}1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{vmatrix}



    For your three vertices:

    . . A \;\;=\;\;\frac{1}{2}\begin{vmatrix}1&0&0 \\ 1 & 14 & 6 \\ 1 & 12 & 9 \end{vmatrix} \;\;=\;\;\frac{1}{2}\left[1\cdot\begin{vmatrix}14 & 6 \\ 12 & 9\end{vmatrix} \:-\: 0\cdot\begin{vmatrix}1&6 \\ 1&9\end{vmatrix} \:+\: 0\begin{vmatrix}1&14 \\ 1&12\end{vmatrix}\,\right]

    . . . . =\;\;\frac{1}{2}\bigg[1(14\cdot9 - 12\cdot6) - 0 + 0\bigg] \;\;=\;\;\frac{1}{2}(126-72) \;\;=\;\;\boxed{27}


    Note: The "determinant bars" also represent absolute value.
    . . . . .The determinent can turn out to be negative.


    Edit: correct a silly oversight . . .
    .
    Last edited by Soroban; March 25th 2010 at 08:49 PM.
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  3. #3
    Member rtblue's Avatar
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    Thanks, I understand now. (although you did forget to divide by 2 at the end )
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