Answer:
3 1/2 square units
Explanation:
For this small triangle whose vertices are on grid points, the use of Pick's theorem will give the area with the least work.
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theorem
Pick's theorem says that the area is ...
A = i +b/2 -1
where i is the number of grid points inside the boundary, and b is the number of grid points on the boundary.
application
Here, only the given coordinates are boundary points that are on grid points. (b = 3)
The figure shows us three grid intersections inside the boundary of the triangle. (i = 3)
That means the area is ...
A = 3 +3/2 -1 = 3 1/2 . . . . square units
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Alternate solution
The triangle is bounded by a square 3 units on a side. From that, we can subtract a 2×3 triangle at upper left, a 1×2 triangle at upper right, and a 3×1 triangle at the bottom. That leave a figure area of ...
A = 3² -1/2(2×3) -1/2(1×2) -1/2(3×1) = 9 -3 -1 -3/2 = 3 1/2 . . . . square units
(The area of a triangle is 1/2bh, where b and h are the base and height of the triangle.)
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Another alternate solution can be had by using a formula with the vertex coordinates:
A = 1/2|(x1)(y2 -y3) +(x2)(y3 -y1) +(x3)(y1 -y2)|
= 1/2|(-8)(3 -5) +(-5)(5 -2) +(-6)(2 -3)| = 1/2|16 -15 +6| = 3 1/2 . . . square units