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

1

The correct answer is Option (1) → 1

Vertices of parallelogram: A(3,2), B(1,−1), C(2,1)

Area of parallelogram = area of triangle ABC × 2.

Using determinant formula for triangle ABC:

$\text{Area of }\triangle ABC =\frac{1}{2}\left| \begin{vmatrix} 3 & 2 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & 1 \end{vmatrix} \right|$

Compute determinant:

$=3\begin{vmatrix}-1 & 1 \\ 1 & 1\end{vmatrix} - 2\begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} + 1\begin{vmatrix}1 & -1 \\ 2 & 1\end{vmatrix}$

$=3((-1)(1) - 1\cdot 1) - 2(1\cdot 1 - 1\cdot 2) + (1\cdot 1 - (-1)\cdot 2)$

$=3(-1 - 1) - 2(1 - 2) + (1 + 2)$

$=3(-2) - 2(-1) + 3$

$= -6 + 2 + 3 = -1$

Area of triangle = $\frac{1}{2}\cdot | -1 | = \frac{1}{2}$

Area of parallelogram = $2 \times \frac{1}{2} = 1$

The area of the parallelogram is 1 square unit.