If A= (2, 3), B=(-1, 0), C= (4, 6) then area of the parallelogram ABCD is :

3 $\frac{3}{2}$ 6 2

3

The correct answer is Option (1) → 3 $\vec{AB}=-3\hat i-3\hat j$ $\vec{BC}=5\hat i+6\hat j$ area = $|\vec{AB}×\vec{BC}|$ so $\vec{AB}×\vec{BC}=\begin{vmatrix}\hat i&\hat j&\hat k\\-3&-3&0\\5&6&0\end{vmatrix}$ $\vec{AB}×\vec{BC}=-3\hat k$ area = 3 sq. units