NIMCET 2012 Question Paper with solutions

We know that,
Centroid of the triangle,
G = $\left(\frac{{\mathit{x}}_1+{\mathit{x}}_2+{\mathit{x}}_3}{3},\frac{{\mathit{y}}_1+{\mathit{y}}_2+{\mathit{y}}_3}{3}\right)$ = (1 , 4)
⇒ $\left\{\frac{4-9+\mathit{x}}{3},\frac{-3+7+\mathit{y}}{3}\right\}$ = (1 , 4)

⇒ $\left(\frac{\mathit{x}-5}{3},\frac{\mathit{y}+4}{3}\right)$ = (1 , 4)
⇒ x - 5 = 3 ⇒ x = 8
and y + 4 = 12 ⇒ y = 8
So, third vertex of a Δ ABC is (8, 8).
Now, area of Δ ABC = $\frac{1}{2}|\begin{array}{ccc}4& -3& 1\\ -9& 7& 1\\ 8& 8& 1\end{array}|$
Use ${\mathit{R}}_2$ → ${\mathit{R}}_2-{\mathit{R}}_1$ , ${\mathit{R}}_3$ → ${\mathit{R}}_3-{\mathit{R}}_1$
= $\frac{1}{2}|\begin{array}{ccc}4& -3& 1\\ -13& 10& 0\\ 4& 11& 0\end{array}|$
Expand with respect to ${\mathit{C}}_3$
= $\frac{1}{3}$ |(- 143 - 40)| = $\frac{1}{2}$ |- 183| = $\frac{183}{2}$