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Find the area of the quadrilaterals, the coordinates of whose vertices are$(-3, 2), (5, 4), (7, -6)$ and $(-5, -4)$
Given:
Vertices of a quadrilateral are $(-3, 2), (5, 4), (7, -6)$ and $(-5, -4)$.
To do:
We have to find the area of the quadrilateral.
Solution:
Let $A(-3, 2), B(5, 4), C(7, -6)$ and $D(-5, -4)$ be the vertices of a quadrilateral $ABCD$.
Join $A$ and $C$ to get two triangles $ABC$ and $ADC$.
This implies,
Area of quadrilateral $ABCD=$ Area of triangle $ABC+$ Area of triangle $ADC$.
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 \( ABC=\frac{1}{2}[-3(4+6)+5(-6-2)+7(2-4)] \)
\( =\frac{1}{2}[-3(10)+5(-8)+7(-2)] \)
\( =\frac{1}{2}[-30-40-14] \)
\( =\frac{1}{2} \times (-84) \)
\( =42 \) sq. units.
Area of triangle \( ADC=\frac{1}{2}[-3(-6+4)+7(-4-2)+-5(2+6)] \)
\( =\frac{1}{2}[-3(-2)+7(-6)+(-5)(8)] \)
\( =\frac{1}{2}[6-42-40] \)
\( =\frac{1}{2} \times (-76) \)
\( =38 \) sq. units.
Therefore,
The area of the quadrilateral $ABCD=42+38=80$ sq. units.
The area of the given quadrilateral is $80$ sq. units.
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