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If A (3, 2), B (1, -1) and C (2, 1) are three vertices of a parallelograms ABCD, then its area (in sq.units) is equal to

Solution

✅ Correct Option: 1

Given three vertices of parallelogram ABCD:

A(3, 2)

B(1, -1)

C(2, 1)

The area of a parallelogram can be found using any three consecutive vertices by calculating the cross product of two adjacent sides.


Vector AB represents the side from A to B:

Vector AB = B - A

Vector AB = (1 - 3, -1 - 2)

Vector AB = (-2, -3)


Vector AC represents the side from A to C:

Vector AC = C - A

Vector AC = (2 - 3, 1 - 2)

Vector AC = (-1, -1)


For a parallelogram with adjacent sides represented by vectors (x₁, y₁) and (x₂, y₂):

Area = ∣x1×y2−x2×y1∣|x₁ \times y₂ - x₂ \times y₁|

Substituting AB = (-2, -3) and AC = (-1, -1):

Area = ∣(−2)×(−1)−(−1)×(−3)∣|(-2) \times (-1) - (-1) \times (-3)|

Area = ∣2−3∣|2 - 3|

Area = ∣−1∣|-1|

Area = 11

Therefore, the area of parallelogram ABCD is 1 square unit.

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