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 $\begin{vmatrix}x_{1}&y_{1}&4\\ x_{2}&y_{2}&4\\ x_{3}&y_{3}&4\end{vmatrix}^{2}$ is

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

Answer 1

Answer: (C)

Given $k = \frac{1}{2} \begin{vmatrix}x_{1}&y_{1}&1\\ x_{2}&y_{2}&1\\ x_{3}&y_{3}&1\end{vmatrix}$
$\Rightarrow 4 \times2k = \begin{vmatrix}x_{1}&y_{1}&4\\ x_{2}&y_{2}&4\\ x_{3}&y_{3}&4\end{vmatrix}$
$\Rightarrow 64 k^{2} = \begin{vmatrix}x_{1}&y_{1}&4\\ x_{2}&y_{2}&4\\ x_{3}&y_{3}&4\end{vmatrix}^{2}$