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Thread: Area triangle using determinant

  1. #1
    Oct 2009

    Area triangle using determinant

    In the accompanying figure ,the area of the triangle $\displaystyle ABC$ can be expressed as

    Area $\displaystyle ABC$ = Area$\displaystyle ADEC$ + Area$\displaystyle CEFB$ - Area$\displaystyle ADFB$
    Use this and the fact that the area of trapezoid equal of the altitude times the sum of the parallel sides to show that

    $\displaystyle Area ABC=\frac{1}{2}\left(\begin{matrix} x_1 &y_2 &1\\x_2 &y_2 & 1 \\x_3 &y_3&1\end{matrix}\right)$

    [Note-in the derivation of this formula, the vertices are labelled such that the triangle is traced counter-clockwise proceeding from $\displaystyle (x_1,y_1)$ to $\displaystyle (x_2,y_2)$ to$\displaystyle (x_3,y_3)$ .for a clockwise orientation, the determinant above yields the negative of the area.]
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  2. #2
    Senior Member
    Jan 2009

    Re: Area triangle using determinant

    $\displaystyle \frac{1}{2}\left(\begin{matrix} x_1 &y_2 &1\\x_2 &y_2 & 1 \\x_3 &y_3&1\end{matrix}\right)$

    You could always expand this, re-arrange and see if you get a form that is equal to Area ADEC + Area CEFB - Area ADFB. In that case, you are done...
    Last edited by MacstersUndead; Nov 14th 2012 at 01:05 PM.
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