Area (in sq. units) of the quadrilateral ABCD with vertices A(2, -1), B(4, 3), (-1, 2), D(-3, -2) is:

Area of quadrilateral ABCD= Area of $$\triangle\ $$DAB+ Area of $$\triangle\ $$BCD

Area of $$\triangle\ $$DAB=$$\ \frac{1 }{ 2}$$*[$$\ -\ $$3($$\ -\ $$1$$\ -\ $$3)+2(3$$\ -\ $$($$\ -\ $$2))+4($$\ -\ $$2$$\ -\ $$($$\ -\ $$1))]

=$$\ \frac{1}{ 2}$$[12$$\ \ +\ $$10$$\ -\ $$4]

=9sq.units

Area of $$\ \triangle\ $$BCD= $$\ \frac{1 }{ 2}$$*[4(2$$\ -\ $$($$\ -\ $$2))+($$\ -\ $$1)($$\ -\ $$2$$\ -\ $$3)+($$\ -\ $$3)(3$$\ -\ $$2)]

=$$\ \frac{1 }{2 }$$*[16$$\ +\ $$5$$\ -\ $$3]=9sq.units

Area of quadrilateralABCD=9$$\ +\ $$9=18sq.units