Find the area of triangle with vertices at points A (1, 1) ,B ( 6, 0) and C ( 3, 2).

Concept:

If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then, 

Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

Note: Area is always a positive quantity, therefore we always take the absolute value of the determinant for the area. 

Calculation:

Given vertices are A (1, 1) ,B ( 6, 0) and C ( 3, 2).

We know that area of triangle ABC is given by, 

Δ = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

⇒ Δ = \(\rm\frac{1}{2}\begin{vmatrix} 1& 1 & 1\\ 6& 0 & 1\\ 3& 2 & 1 \end{vmatrix}\) 

⇒ Δ = \(\frac{1}{2}\left [ 1\left ( 0-2 \right )-1 (6-3)+1 (12-0)\right ]\) 

⇒ Δ =  \(\frac{7}{2}\)  

The correct option is 1.