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If $P (-5, -3), Q (-4, -6), R (2, -3)$ and $S (1, 2)$ are the vertices of a quadrilateral $PQRS$, find its area.
Given:
$P (-5, -3), Q (-4, -6), R (2, -3)$ and $S (1, 2)$ are the vertices of a quadrilateral $PQRS$.
To do:
We have to find the area of the quadrilateral.
Solution:
Join $P$ and $R$ to get two triangles $PQR$ and $PSR$.
This implies,
Area of quadrilateral $PQRS=$ Area of triangle $PQR+$ Area of triangle $PSR$.
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( PQR=\frac{1}{2}[-5(-6+3)+(-4)(-3+3)+2(-3+6)] \)
\( =\frac{1}{2}[-5(-3)+(-4)(0)+2(3)] \)
\( =\frac{1}{2}[15+0+6] \)
\( =\frac{1}{2} \times (21) \)
\( =\frac{21}{2} \) sq. units.
Area of triangle \( PSR=\frac{1}{2}[-5(2+3)+1(-3+3)+2(-3-2)] \)
\( =\frac{1}{2}[-5(5)+1(0)+2(-5)] \)
\( =\frac{1}{2}[-25+0-10] \)
\( =\frac{1}{2} \times (-35) \)
\( =\frac{35}{2} \) sq. units.
Therefore,
The area of the quadrilateral $PQRS=\frac{21}{2}+\frac{35}{2}=\frac{21+35}{2}=28$ sq. units.
The area of the quadrilateral $PQRS$ is $28$ sq. units.
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