# Area triangle using determinant

• Nov 13th 2012, 11:38 PM
mastermin346
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 ½ 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.]
• Nov 14th 2012, 01:00 PM
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...