
To find the area of triangle whose vertices are given. Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of the triangle. From A,B,C draw the perpendiculars AL, BM and CN on the x axis. Area of triangle ABC = area of trap ABML + trap ALNC  trap BMNC = 1/2[MB+LA]ML + 1/2[LA+CN]LN  1/2[MB+NC]MN = 1/2(y_{2}+y_{1}) (x_{1}x_{2}) + 1/2(y_{1}+y_{3})(x_{3}x_{1})  1/2(y_{2}+y_{3})(x_{3}x_{2}) = 1/2[x_{1}(y_{2}y_{3})+x_{2}(y_{3}y_{1})+x_{3}(y_{1}y_{2})] Alternatively 1/2 Example: Find the area of triangle whose vertices are (1,3), (2,4) and (5,6)
1/2[1x42x3+2x65x4+5x31x6] = 1/2 Changing the sign, the area of the triangle in magnitude is 1/2.
Directions: Solve the following. 