Question

The area of the parallelogram with vertices \( (0, 0), (7, 2), (5, 9) \) and \( (12, 11) \) is:

• A. \( 50 \)
• B. \( 54 \)
• C. \( 51 \)
• D. \( 52 \)
• E. \( 53 \)

Hint:

For any parallelogram \( ABCD \), the area is twice the area of triangle \( ABC \). Using the determinant method directly on the vectors from the origin is the fastest way to solve coordinate-based area problems.

Answer 1

Answer: (E)

Concept: The area of a parallelogram with one vertex at the origin \( (0,0) \) and adjacent vertices at \( (x_1, y_1) \) and \( (x_2, y_2) \) is the absolute value of the determinant of the matrix formed by those two vectors: \( \text{Area} = |x_1y_2 - x_2y_1| \).

Step 1: Identify adjacent vectors from the origin.
The vertices are \( O(0,0), A(7,2), B(5,9) \). The fourth vertex \( (12,11) \) is simply the vector sum \( A+B \), confirming \( A \) and \( B \) are adjacent to the origin. Adjacent vectors: \( \vec{u} = (7, 2) \) and \( \vec{v} = (5, 9) \).

Step 2: Calculate the determinant.
\[ \text{Area} = |(7)(9) - (2)(5)| \] \[ \text{Area} = |63 - 10| = 53 \]