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On a coordinate plane, 2 parallelograms are shown. Parallelogram 1 has points (0, 2), (2, 6), (6, 4), and (4, 0). Parallelogram 2 has points (2, 0), (4, negative 6), (2, negative 8), and (0, negative 2).
How do the areas of the parallelograms compare?
The area of parallelogram 1 is 4 square units greater than the area of parallelogram 2.
The area of parallelogram 1 is 2 square units greater than the area of parallelogram 2.
The area of parallelogram 1 is equal to the area of parallelogram 2.
The area of parallelogram 1 is 2 square units less than the area of parallelogram 2.
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GPT-5 mini
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Compute each parallelogram's area as the magnitude of the cross product of two adjacent side vectors.
Parallelogram 1: use (0,2) as a vertex. Side vectors v = (2,6) − (0,2) = (2,4) and w = (4,0) − (0,2) = (4,−2).
Area1 = |det([v,w])| = |2(−2) − 4·4| = |−4 − 16| = 20.
Parallelogram 2: use (2,0). Side vectors v = (4,−6) − (2,0) = (2,−6) and w = (0,−2) − (2,0) = (−2,−2).
Area2 = |det([v,w])| = |2(−2) − (−6)(−2)| = |−4 − 12| = 16.
Thus parallelogram 1 has area 4 square units greater than parallelogram 2.
Parallelogram 1: use (0,2) as a vertex. Side vectors v = (2,6) − (0,2) = (2,4) and w = (4,0) − (0,2) = (4,−2).
Area1 = |det([v,w])| = |2(−2) − 4·4| = |−4 − 16| = 20.
Parallelogram 2: use (2,0). Side vectors v = (4,−6) − (2,0) = (2,−6) and w = (0,−2) − (2,0) = (−2,−2).
Area2 = |det([v,w])| = |2(−2) − (−6)(−2)| = |−4 − 12| = 16.
Thus parallelogram 1 has area 4 square units greater than parallelogram 2.
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