(-1,3) (2,0) (2,-1) (-5,-1)
The easiest way to go about computing the area of a general convex quadrilateral is to divide it into two triangles by constructing a line connecting two opposite vertices, computing each of the triangle areas, and then summing these two areas. I selected the (-1,3) and (2,-1) vertices because it resulted in each of the resulting triangles having a base parallel to one of the coordinate axes, thus simplifying the calculations significantly.
The green lines are the triangle altitudes. Next you need the lengths of the two triangle bases and the two altitudes. You can use the distance formula, but since all four of these lines are parallel to the axes, the distance formula degenerates to the magnitude of the difference of the unequal coordinates.
So the area of triangle ABD is:
And:
So the area of triangle BCD is:
And finally, the area of quadrilateral ABCD is:
This problem could have been done by constructing the line segment AC, but then to determine the length of the base of triangle ACD, you would have to use the distance formula to calculate the distance from A to C:
Then you would have to use the Two-Point form to derive the equation of the line containing segment AC, and in the process calculate the slope of the line containing the segment AC,
calculate the negative reciprocal of that slope:
where point G is the point of intersection between AC and a perpendicular to AC passing through D. Using this slope, you would then determine the equation of that perpendicular with the point-slope form:
Then you would have to solve the system of the two equations, line containing AC and line containing DG to obtain the coordinates of point G.
Next, you would have to calculate the length of segment DG using the distance formula:
And at last you would have the lengths of the base and altitude for triangle ACD so that you could compute the area:
Which you would then have to add to the area of triangle ABC.
See what I mean about significantly simpler calculations?
John