If P(–5, –3), Q(–4, –6), R(2, –3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.

#### Solution

The vertices of the quadrilateral PQRS are P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2).

Join QS to form two triangles, namely, ΔPQS and ΔRSQ.

Area of quadrilateral PQRS = Area of ΔPQS + Area of ΔRSQ

We know

Area of triangle having vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) = `1/2`[x_{1}(y_{2}−y_{3})+x_{2}(y_{3}−y_{1})+x_{3}(y_{1}−y_{2})]

Now,

Area of △PQS = `1/2`[−5(−6−2)+(−4)(2+3)+1(−3+6)]

=`1/2`[−5(−8)+(−4)(5)+1(3)]

=`1/2`(40−20+3)

=`23/2`square units

Area of ∆RSQ= `1/2`[2(2+6)+1(−6+3)+(−4)(−3−2)]

=`1/2`[2(8)+1(−3)+(−4)(−5)]

=`1/2`(16−3+20)

=`33/2` square units

Area of quadrilateral PQRS = Area of ΔPQS + Area of ΔRSQ

`=23/2+33/2`

`=56/2`

=28 square units

Thus, the area of quadrilateral PQRS is 28 square units.