Question

On a coordinate plane, triangle A B C and parallelogram G H J K are shown. Triangle A B C has points (2, 0), (1, negative 6), (negative 2, negative 4). Parallelogram G H J K has points (0, 0), (1, 2), (negative 2, 4), and (negative 3, 2).
How does the area of triangle ABC compare to the area of parallelogram GHJK?

The area of △ABC is 2 square units greater than the area of parallelogram GHJK.
The area of △ABC is 1 square unit greater than the area of parallelogram GHJK.
The area of △ABC is equal to the area of parallelogram GHJK.
The area of △ABC is 1 square unit less than the area of parallelogram GHJK.

Answer 1

Compute each area.

Triangle ABC with A(2,0), B(1,-6), C(-2,-4):
Area = 1/2 |x1y2 + x2y3 + x3y1 − (y1x2 + y2x3 + y3x1)|
= 1/2 |2(−6) + 1(−4) + (−2)(0) − (0·1 + (−6)(−2) + (−4)·2)|
= 1/2 |−12 −4 + 0 − (0 +12 −8)| = 1/2 |−16 −4| = 1/2 ·20 = 10.

Parallelogram GHJK with G(0,0), H(1,2), K(−3,2):
Two adjacent side vectors from G are GH = (1,2) and GK = (−3,2), so area = |det([1 2],[−3 2])| = |1·2 − 2·(−3)| = 8.

Thus triangle area = 10, parallelogram area = 8, so the triangle is 2 square units greater.

Answer: The area of △ABC is 2 square units greater than the area of parallelogram GHJK.