The area of triangle with vertices A(x1,y1),B(x2,y2) and C(x3,y3) is:
Let ABC be any triangle whose vertices are A(x1,y1),B(x2,y2)and C(x3,y3). Draw AP, BQ and CR perpendiculars from A, B and C, respectively, to the x-axis. Clearly ABQP, APRC and BQRC are all trapeziums.
Now from the figure, it is clear that
Area of △ABC = Area of trapezium ABQP + Area of trapezium APRC - Area of trapezium BQRC.
Area of a trapezium =12 (Sum of parallel sides)(Distance between the parallel sides)
∴ Area of ΔABC=12(BQ+AP)QP+12(AP+CR)PR−12(BQ+CR)QR
=12(y2+y1)(x1−x2)+12(y3+y1)(x3−x1)−12(y2+y3)(x3−x2)
=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
Thus, the area of ΔABC is the numerical value of the expression 12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]