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Q.
If the vertices of quadrilateral are $(-4,5),(0,7)$, $(5,-5)$ and $(-4,-2)$, then area of quadrilateral is
Straight Lines
Solution:
Given vertices $A(-4,5), B(0,7), C(5,-5)$ and $D(-4,-2)$ are shown below in the $x y$-plane. Joining all the vertices a quadrilateral $A B C D$ is formed.
Area of quadrilateral $A B C D$
$=$ Area of $\triangle A D C+$ Area of $\triangle A B C .....$(i)
Now, area of $\triangle A D C$
$=\frac{1}{2}|[-4(-2+5)-4(-5-5)+5(5+2)]$
$[\because $ Area $=\frac{1}{2} |x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)$ and here,
$\left(x_1, y_1\right)=(-4,5),\left(x_2, y_2\right)=(-4,-2),\left(x_3, y_3\right)=(5,-5)]$
$=\frac{1}{2}|[-4 \times 3-4(-10)+5 \times 7]|$
$=-\frac{1}{2}|[-12+40+35]|=\frac{1}{2}|[75-12]|=\frac{63}{2} $
Area of $ \triangle A B C=\frac{1}{2}|[-4(7+5)+0(-5-5)+5(5-7)]| $
${\left[\text { here, }\left(x_1, y_1\right)\right.}\left.=(-4,5),\left(x_2, y_2\right)=(0,7) \operatorname{and}\left(x_3, y_3\right)=(5,-5)\right] $
$\frac{1}{2}|[-4 \times 12+0 \times(-10)+5(-2)]| $
$=\frac{1}{2}|-48-10| $
$=\frac{|-58|}{2}=29$
$ \therefore $ From Eq. (i)
Area of quadrilateral $ A B C D$
$=\frac{63}{2}+29=\frac{63+58}{2}=\frac{121}{2} $
$=60.5 $ sq units
