Question

Find the area of a quadrilateral $ABCD$ whose vertices area $A(3,-1),B(9,-5),C(14,0),D(9,19)$.

Answer 1

Vertices of quadrilateral are

$A(3,-1)B(9,-5)C(14,0)D(9,19)$

We have $A(x_1,y_1)=(3,-1)$

$B(x_2,y_2)=(9,-5)$

$C(x_3,y_3)=(14,0)$

$D\left(\right(x_4,y_4)=(9,19)$

now area of quadrilateral $ABCD=$ [Ref. image]

$=\frac{1}{2}\left[\right(x_1{y}_2+x_2{y}_3+x_3{y}_4+x_4{y}_1)-(x_2{y}_1+x_3{y}_2+x_4{y}_3+x_1{y}_4\left)\right]$

$=\frac{1}{2}\left[\right(-15+0+(14\times 19)-19]-(-9-70+0+57\left)\right]$

$=\frac{1}{2}\left[\right(-15-19+266\left)\right]-(-79+57)]$

$=\frac{1}{2}[266-34+22]$

$=\frac{1}{2}[266-12]=\frac{1}{2}\left[254\right]$

Ar. $ABCD=127sq.unit$

Hence area of quadrilateral $ABCD$ is $127sq.unit$