Find the area of the parallelogram with vertices A(−3, 0), B(−1, 4), C(6, 3), and D(4, −1).?

Solution:

The vertices of a parallelogram are A(−3, 0), B(−1, 4), C(6, 3), and D(4, −1)

Area of triangle ABC = \( \frac{1}{2}\begin{vmatrix} -3 & 0 & 1\\ -1 & 4 & 1\\ 6 & 3 & 1 \end{vmatrix} \)

= 1/2 [-3 (4 - 3) - 1 (- 3 - 24)]

= 1/2 [-3 (1) - 1 (-27)]

= 1/2 [-3 + 27]

= 1/2 [24]

= 12 sq units

Area of triangle ACD = \( \frac{1}{2}\begin{vmatrix} -3 & 0 & 1\\ 6 & 3 & 1\\ 4 & -1 & 1 \end{vmatrix}\)

= 1/2 [-3 (3 + 1) - 1 (- 6 - 12)]

= 1/2 [-3 (4) - 1 (-18)]

= 1/2 [-12 + 18]

= 1/2 [6]

= 3 sq units

Area of parallelogram ABCD = Area of triangle ABC + Area of triangle ACD

= 12 + 3

= 15 sq units.

Therefore, the area of the parallelogram is 15 sq. units.

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Find the area of the parallelogram with vertices A(−3, 0), B(−1, 4), C(6, 3), and D(4, −1).?

Summary:

The area of the parallelogram with vertices A(−3, 0), B(−1, 4), C(6, 3), and D(4, −1) is 15 sq. units.