Find the area of the quadrilateral PQRS the coordinates of whose vertices are P(–4, –2), Q(–3, –5), R(3, –2) and S(2, 3).

Given: P(– 4, –2), Q(–3, –5), R(3, –2) and S(2, 3)

By joining Q to S, we will get two triangles PQS and QRS.

So, Area of quadrilateral PQRS = Area of ΔPQS + Area of ΔQRS

We know that the area of Triangle = .

Now area of ΔPQS = [(-4)(-5 – 3) + (-3)(3 – (-2)) + 2(-2 – (-5))] / 2

= [(-4)(-8) + (-3)(5) + 2(-3)] / 2

= [32 – 15 - 6] / 2

=11 / 2

∴ Area of ΔABD = 11/2 sq. Units

Also, area of ΔBCD = [(-3)(-2 – 3) + 3(3 – (-5)) + 2(-5 – (-2))] / 2

= [-3(-5) + 3(8) + 2(-3)] / 2

= [-15 + 24 – 6] / 2

= 3/2

∴ Area of ΔBCD = 3/2 sq units

So, Area of a quadrilateral = 1/2 + 3/2 = 4/2 = 2 sq. units

Ans. The area of quadrilateral is 2 sq. units.