How-to find the area of the Triangle - ∆

Can be found by the below Square Matrix. Determinant value of the Matrix yields the surface of the triangle ∆. And it can be denoted by the symbol delta as well - δ

|x1 y1 1|
|x2 y2 1| . 1/2
|x3 y3 1|



Let AL, BM, CN be the ordinates of the points A, B and C respectively.

Now  ∆ ABC = trapezium ABML + trap. ALNC - trap BMNC
                      = 1/2(AL+BM)ML+1/2(AL+CN)LN-1/2(BM+CN)MN
=>              ∆ = 1/2 (y1+y2)(x1-x2)+1/2(y1+y3)(x3-x1)-1/2(y2+y3)(x3-x2)
                      = 1/2[x1y2-x2y1+x3y1-x1y3+x2y3-x3y2]
∴              ∆   = 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
i.e.,           ∆   = 1/2Σx1(y2-y3) numerically.

Note:
        |x1 y1 1|
1. ∆=|x2 y2 1| . 1/2
        |x3 y3 1|
2. The area obtained by the formula is positive or negative according as the vertices are taken round the triangle anti-clockwise or clockwise
3. If the area of triangle formed by the three points is zero, then the points are collinear,
                                                             |x1 y1 1|
i.e., if ∆=0, then 1/2.Σx1(y2-y3) = 0 or |x2 y2 1| =0
                                                             |x3 y3 1|
is the condition for collinearity of three points.



Source: Polytechnic Mathematics-2, S.Janakiraman, J.J Publications