If A (3, 2), B (1, −1) and C (2, 1) are three vertices of a parallelogram ABCD, then its area (in sq. units) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
The diagonal AC divides the parallelogram into two equal triangles, so the area of ABCD is twice the area of triangle ABC.
\( \displaystyle \text{Area of }\triangle ABC = \frac{1}{2}\Big|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\Big| \)
\( \displaystyle = \frac{1}{2}\Big|3(-1 - 1) + 1(1 - 2) + 2(2 + 1)\Big| \)
\( \displaystyle = \frac{1}{2}\Big|-6 - 1 + 6\Big| = \frac{1}{2} \)
\( \displaystyle \text{Area of } ABCD = 2\times\frac{1}{2} = 1 \) sq. unit
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{2}\) — Incorrect. This is the area of triangle ABC only, which is half the parallelogram.
Option (3) 5 — Incorrect. The area is \(2\times\frac{1}{2} = 1\), not 5.
Option (4) \(\displaystyle\frac{5}{2}\) — Incorrect. The area is 1, not \(\frac{5}{2}\).