Question

If A( -3, 5), B(- 2, -7), C(1, - 8) and D(6, 3) are the vertices of a quadrilateral ABCD then find its area.

Answer 1

Given, vertices of quadrilateral : $A(-3,5),B(-2,-7),C(1,-8)andD(6,3)$

Area of quadrilateral = Area of $△ABC$ + Area of $△DAC$

Area of triangle = $\frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)$

Area of $△ABC$ = $\frac{1}{2}|(-3(-7+8)-2(-8-5)+1(5+7\left)\right)|$

= $\frac{1}{2}|(-3+26+12)|$

= $\frac{35}{2}$
Area of $△ADC$ = $\frac{1}{2}|(-3(3+8)+6(-8-5)+1(5-3\left)\right)|$

= $\frac{1}{2}|(-33-78+2)|$

= $\frac{109}{2}$
Area of quadrilateral = $\frac{35}{2}+\frac{109}{2}$

Area of quadrilateral = $72c{m}^2$