Answer:
(2) 11
Explanation:
You want the area of the triangle with vertices (1, 3), (3, -1), and (-2, -2).
Pick's theorem
Pick's theorem says the area of a figure drawn on a grid can be found by counting the grid points on the boundary (b) and the interior grid points (i). Then the area is ...
A = (b/2) +i -1
This figure has a boundary grid point at (2, 1) in addition to the three vertices. So, b = 4.
There are 10 interior grid points, a number that can be arrived at by counting them. The numbers in each row, starting from the bottom, are 4, 3, 2, 1.
Then the area is ...
A = (4/2) +10 -1 = 11
The area of the triangle is 11 square units.
From Coordinates
The area can be computed from the coordinates by the formula ...
A = 1/2|x1(y2 -y3) +x2(y3 -y1) +x3(y1 -y2)|
A = 1/2|1(-1 -(-2)) +3(-2 -3) +(-2)(3 -(-1))| = 1/2|1 -15 -8| = 11
The area is 11 square units.
Using geometry
The triangle can be bounded by a square 5 units on a side. The triangle area is smaller than the square area by the sum of the three triangles cut from the corners of the square. Clockwise from left, those triangle areas are ...
1/2(5·3) +1/2(4·2) +1/2(1·5) = 1/2(15 +8 +5) = 14
Then the area of the triangle of interest is ...
5² -14 = 11
The area of the triangle is 11 square units.
Using trigonometry
The angle the bottom edge makes with the horizontal is ...
arctan(1/5) ≈ 11.3099°
and the length of that edge is ...
5/cos(11.3099°) ≈ 5.099
The angle the left side makes with the horizontal is ...
arctan(5/3) ≈ 59.0362°
and the length of that edge is ...
3/cos(59.0362°) ≈ 5.831
The area of the triangle with sides 'a' and 'b' and angle C between them is ...
Area = 1/2(ab·sin(C)) = 1/2(5.099·5.831·sin(59.0362° -11.3099°)) = 11
The area of the triangle is 11 square units.
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Additional comment
Side lengths could also be determined using the Pythagorean theorem, and the angle between any two of them could be found using the Law of Cosines.
Heron's formula could be used to find the area from the three side lengths.