A(-12,2) , B(6,2),C(-2,-3) and D(-20,-3)
Let a digonal BD is drawn . Then we get two triangles
Delta DAB and Delta DCB ; for Delta DAB ,
D(-20,-3),A(-12,2) , B(6,2) Area Delta DAB =
A_(DAB) = |1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|
=|1/2(-20(2-2)+(-12(2+3)+6(-3-2)|
=|1/2(0-60-30)| =|1/2 *(-90)| = | -45| =45
For Delta DCB
D(-20,-3),C(-2,-3) , B(6,2) Area Delta DCB =
A_(DCB) = |1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|
=|1/2(-20(-3-2)+(-2(2+3)+6(-3+3)|
=|1/2(100-10+0)| =|1/2 *(90)| = | 45| =45
Area of parallelogram ABCD = A_(DAB)+A_(DCB)
=45+45 =90 Sq.unit