In the accompanying gure, the area of the triangleABC can be expressed as Cm. 3'3] area ABC = area ADEC + area CEFB - area ADFB 399-le Use this and the fact that the area of a trapezoid equals 5 the altitude times the sum of the parallel sides to show that 1 361 3'1 1 area ABC 2 3 12 322 I3 3'3 1 Note In the derivation of this formula, the vertices are labeled such that the triangle is traced counterclockwise proceeding from (x1 , ml to (X2 , Y2} to (X3 , y3} . For a clockwise orientation, the determinant above yields the negative of the area. (You should not solve (a)) (b) Use the result in (a) to nd the area of the triangle with vertices (-6 , -7} , (9 , O) , (3 , 3}. Area = n