A   B   C   D   A
-3  -2   4   0  -3
 4  -1   4   5   4
--------------
Add the diagonal products starting at the upper left.
--> 3  -8  +20  + 0 = 15
----
Add the diagonal products starting at the lower left.
--> -8 -4 + 0 -15 = -27

I will assume that the vertices of the quadrilateral go in that order as the points are listed.


Then I see that the diagonal AC of the quadrilateral is horizontal y=4 and divides the quadrilateral in two triangles.


Triangle ABC has the vertex  B  BELOW  the diagonal line y= 4. 

It has the base AC of the length 4 - (-3) = 7 units and the altitude drawn to AC of the length 4 - (-1) = 5 units.

Hence, the area of the triangle ABC is   = 17.5 square units.



Next, triangle ADC has the vertex  D  ABOVE  the diagonal line y= 4. 

It has the base AC of the length 4 - (-3) = 7 units and the altitude drawn to AC of the length 5 - 4 = 1 units.

Hence, the area of the triangle ADC is   = 3.5 square units.



Finally, the area of the quadrilateral ABCD is the sum of the areas of the triangles ABC and ADC


     = 17.5 + 3.5 = 21 square units.    ANSWER