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 (x1, y1), (x2, y2) and (x3, y3) = `1/2`[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
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.