There are a number of ways the area can be figured. One of them is to create points b'(-5, 0) and c'(3, 0). Then subtract the areas of triangles bb'a and cc'a from trapezoid bb'c'c. Here is that computation.
(area of bb'c'c) = (1/2)(3 + 2)(3-(-5)) = 20
(area of bb'a) = (1/2)3*5 = 15/2
(area of cc'a) = (1/2)(2*3) = 3
Area of abc = 20 - 15/2 - 3 = 19/2 = 9 1/2 square units.
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The equivalent computation to that above is
area = |bx*cy -by*cx|/2 = |-5*2-3*3|/2 = 19/2 . . . square units
The formula for this derives from considering points pairwise as defining trapezoids with respect to the x-axis. Summing the areas of those proceeding clockwise around the figure will give a simple formula for area from vertices.
