The area of the triangle with vertices (1, 2) , (4, -1) and (-2, 3) is

The correct answer is Option (4) → 3 square units

The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is the absolute value of

    \( \displaystyle \Delta = \frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix} \)

Here

    \( \displaystyle \Delta = \frac{1}{2}\begin{vmatrix}1 & 2 & 1\\ 4 & -1 & 1\\ -2 & 3 & 1\end{vmatrix} \)

Expanding along the first row:

    \( \displaystyle \Delta = \frac{1}{2}\big[1(-1 - 3) - 2(4 + 2) + 1(12 - 2)\big] \)

    \( \displaystyle \Delta = \frac{1}{2}(-4 - 12 + 10) \)

    \(\Delta = -3\)

Area is never negative, so the area is \(|-3| = 3\) square units.