Question

If $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ are the vertices of a triangle whose area is 'k’ square units, then ${\left|\begin{array}{ccc}{x}_1& y_1& 4\\ x_2& y_2& 4\\ x_3& y_3& 4\end{array}\right|}^2$ is

• A. $32/k^2$
• B. $16/k^2$
• C. $64/k^2$
• D. $48/k^2$

Answer 1

Answer: (C)

Given $k=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$
$\Rightarrow 4\times 2k=\left|\begin{array}{ccc}{x}_1& y_1& 4\\ x_2& y_2& 4\\ x_3& y_3& 4\end{array}\right|$
$\Rightarrow 64k^2={\left|\begin{array}{ccc}{x}_1& y_1& 4\\ x_2& y_2& 4\\ x_3& y_3& 4\end{array}\right|}^2$