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(C) The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is given by the formula:$\frac{1}{2} \left|\begin{array

Vedclass · Accepted Answer

$64$

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(C) The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is given by the formula:$\frac{1}{2} \left|\begin{array}{lll}x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \ x_{3} & y_{3} & 1\end{array}ight| = k$This implies that $\left|\begin{array}{lll}x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \ x_{3} & y_{3} & 1\end{array}ight| = 2k$.Now,consider the given determinant:$D = \left|\begin{array}{lll}x_{1} & y_{1} & 4 \ x_{2} & y_{2} & 4 \ x_{3} & y_{3} & 4\end{array}ight|$Taking the constant $4$ common from the third column,we get:$D = 4 \left|\begin{array}{lll}x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \ x_{3} & y_{3} & 1\end{array}ight| = 4(2k) = 8k$.Therefore,the square of the determinant is:$D^{2} = (8k)^{2} = 64k^{2}$.

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 (in $k^{2}$)

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