Let \(\Delta\) be the area of the triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\) and \((x_3, y_3)\). Then \(\begin{vmatrix}x_1 & x_2 & x_3\\ y_1 & y_2 & y_3\\ 1 & 1 & 1\end{vmatrix}^{2}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(4\Delta^{2}\)
The area of a triangle is
\( \displaystyle \Delta = \left|\,\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}\,\right| \)
So the determinant \(D\) in this formula is \(\pm 2\Delta\).
The determinant in the question is the transpose of D, and transposing does not change a determinant. Hence
\((\pm 2\Delta)^{2} = 4\Delta^{2}\)
Check with \((0, 0)\), \((2, 0)\), \((0, 2)\): the area is 2, the determinant is 4, and \(4^{2} = 16 = 4 \times 2^{2}\).