Question

Draw a quadrilateral in the Cartesian plane, whose vertices are $(-4,5),(0,7),(5,-5)$ and $(-4,-2)$. Also, find its area.

Answer 1

$A\left(ABCD\right)=A\left(△ABC\right)+A\left(△ACD\right)$

$(-4,5)\equiv (x_1,y_1),(0,7)\equiv (x_2,y_2),(5,-5)\equiv (x_3,y_3)$

Now, area of $△ABC$

$=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_2)+x_3(y_1-y_2)|$

$=\frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|$

$=\frac{1}{2}|-4(12)+5(-2\left)\right|$

$=\frac{1}{2}|-48-10|$

$=\frac{1}{2}|-58|$

$=\frac{1}{2}\times 58$

$=29$ sq.units

Area of $△ACD$

$=\frac{1}{2}|-4(-5+2)+5(-2-5)+(-4)(5+5)|$

$=\frac{1}{2}|-4(-3)+5(-7)-4(10\left)\right|$

$=\frac{1}{2}|12-35-40|$

$=\frac{1}{2}|-63|$

$=\frac{63}{2}$ sq.units

So, area ($ABCD$)$=(29+\frac{63}{2})$

$=\frac{58+63}{2}$

$=\frac{121}{2}$ sq.units