Question

Find the area of the quadrilateral, the coordinates of whose vertices are
$(-4,-2),(-3,-5),(3,-2),(2,3)$

Answer 1

We know that,

Area of the quadrilateral ABCD =Area of $△ABC$+Area of $△ACD$

Area of $△ABC$

$=\frac{1}{2}\mid y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\mid$ sq.units

Here, $x_1=-4,y_1=-2,x_2=-3,y_2=-5,x_3=3,y_3=-2$

$=\frac{1}{2}\mid -2(-3-3)+(-5)(3+4)-2(-4+3)\mid$ sq.units

$=\frac{1}{2}\mid -2(-6)+(-5)\left(7\right)-2(-1)\mid$ sq.units

$=\frac{1}{2}\mid 12-35+2\mid$ sq.units

$=\frac{1}{2}\mid -21\mid =10.5$ sq.units

Area of $△ACD$

$=\frac{1}{2}\mid y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\mid$ sq.units.

Here, $x_1=-4,y_1=-2,x_2=3,y_2=-2,x_3=2,y_3=3$

$=\frac{1}{2}\mid -2(3-2)+(-2)(2+4)+3(-4-3)\mid$sq.units

$=\frac{1}{2}\mid -2-12-21\mid$sq.units

$=\frac{1}{2}\times 35$ sq.units

$=17.5$ Sq.units

$\therefore$ Area of the quadrilateral ABCD $=10.5+17.5$$=28$ Sq.units