Find the area of quadrilateral PQRS whose vertices are

Question:

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

Solution:

By joining P and R, we get two triangles PQR and PRS.

Let $P\left(x_{1}, y_{1}\right)=P(-5,-3), Q\left(x_{2}, y_{2}\right)=Q(-4,-6), R\left(x_{3}, y_{3}\right)=R(2,-3)$ and $S\left(x_{4}, y_{4}\right)=S(1,2)$. Then

Area of $\Delta P Q R=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$=\frac{1}{2}[-5(-6+3)-4(-3+3)+2(-3+6)]$

$=\frac{1}{2}[15-0+6]=\frac{21}{2}$ sq. units

Area of $\Delta P R S=\frac{1}{2}\left[x_{1}\left(y_{3}-y_{4}\right)+x_{3}\left(y_{4}-y_{1}\right)+x_{4}\left(y_{1}-y_{3}\right)\right]$

$=\frac{1}{2}[-5(-3-2)+2(2+3)+1(-3+3)]$

$=\frac{1}{2}[25+10+0]=\frac{35}{2}$ sq. units

So, the area of the quadrilateral $P Q R S$ is $\frac{21}{2}+\frac{35}{2}=28$ sq. units sq. units.

 

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