Find the area of the parallelogram with vertices a(-3, 0), b(-1, 3), c(5, 2), and d(3, -1)?

Solution:

The vertices of a parallelogram are a(−3, 0), b(−1, 3), c(5, 2), and d(3, −1)

Area of triangle abc = \( \frac{1}{2}\begin{vmatrix} -3 & 0 & 1\\ -1 & 3 & 1\\ 5 & 2 & 1 \end{vmatrix} \)

= 1/2 [-3 (3 - 2) - 1 (- 2 - 15)]

= 1/2 [-3 (1) - 1 (-17)]

= 1/2 [-3 + 17]

= 1/2 [14]

= 7 sq units

Area of triangle acd = \( \frac{1}{2}\begin{vmatrix} -3 & 0 & 1\\ 5 & 2 & 1\\ 3 & -1 & 1 \end{vmatrix}\)

= 1/2 [-3 (2 + 1) - 1 (- 5 - 6)]

= 1/2 [-3 (3) - 1 (-11)]

= 1/2 [-9 + 11]

= 1/2 [2]

= 1 sq units

Area of parallelogram abcd = Area of triangle abc + Area of triangle acd

= 7 + 1

= 8 sq units.

Therefore, the area of the parallelogram is 8 sq. units.

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Find the area of the parallelogram with vertices a(-3, 0), b(-1, 3), c(5, 2), and d(3, -1)?

Summary:

The area of the parallelogram with vertices a(-3, 0), b(-1, 3), c(5, 2), and d(3, -1) is 8 sq. units.