Here, we have x1 = -4, y1 = -2
x2 = -3, y2 = -5
and x3 = 3, y3 = -2
Now. area of ∆ABC
We know, area of ∆ACD
Here, we have x1 = -4, y1 = -2
x2 = 3, y2 = -2
and x3 = 2, y3 = 3
Now, area of ∆ACD
Hence, the area of quadrilateral
ABCD = ar (∆ABC) + ar (∆ACD)
= (10.5 + 17.5) = 28 sq. units.