Here is one way.

Since the coordinates of the corners are given, then we can get the lengthjs of the 3 sides

distance = sqrt[(x2 -x1)^2 +(y2 -y1)^2]

Then we can get the cos (angle ACB) by using the Law of Cosines.

(AB)^2 = (AC)^2 +(BC)^2 -2(AC)(BC)*cos(angle ACB)

cos(angle ACB) = [(AC)^2 +(BC)^2 -(AB)^2] / [2(AC)(BC)] ---(1)

Then we can get the area of triangle ABC by using the Heron's Formula.

A = sqrt[s(s-a)(s-b)(s-c)]

Or, by using the formula, A = (1/2)bc*sinA

Those above need long computations. I don't need to show those computations here, so solve them yourself.

I solved them on paper here and I got:

AB = sqrt(18)

BC = sqrt(5)

AC = sqrt(5)

Substituting those in (1), I got cos(angle ACB) = -8/10 = -4/5.

Then using the formula A = (1/2)bc*sinA for the area of triangle A,

where

angle "A" = angle ACB

"b" = AC = sqrt(5)

"c" = BC = sqrt(5)

cos(A) = -4/5

Square both sides,

cos^2(A) = 16/25

So,

1 -sin^2(A) = 16/25

sin^2(A) = 1 -16/25 = 9/25

sinA = sqrt(9/25) = 3/5

So, area of triangle ABC

= (1/2)*sqrt(5)*sqrt(5)*(3/5)

= (1/2)(5)(3/5)

= 3/2 or 1.5 sq. units ----------answer.