Find the area of the quadrilateral whose vertices taken in order are (-4,-2), (-3,-5), (3,-2), and (2,3).

Let A(-4,-2), B(-3,-5), C(3,-2), and D(2,3) be the vertices of the quadrilateral ABCD.
Area of a quadrilateral ABCD = Area of △ABC + Area of △ACD
By using a formula for the area of a triangle = 1/2|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
Area of △ABC = 1/2[-4(-5+2) + (-3)(-2+2) + 3(-2+5)] = 1/2[12 + 0 + 9] = 21/2 sq. units
Area of △ACD = 1/2[-4(3+2) + 3(-2+2) + 2(-2-3)] = 1/2[-20 + 0 -10] = -30/2 = -15 sq.units
However, area cannot be negative, so we take the absolute value: 15 sq. units
Therefore, Area of quadrilateral = 21/2 + 15 = (21 + 30)/2 = 51/2 = 25.5 sq. units