The area of a triangle whose vertices are $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the absolute value of

The area of a triangle with vertices at $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:

$\text{Area} = \frac{1}{2}\left| \begin{array}{ccc} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{array} \right|$

The formula requires three essential components:

The factor $\frac{1}{2}$ ensures the correct area is calculated. Without it, the result would be double the actual area.

All three entries in the last column must be 1. The pattern for each vertex is: (x-coordinate, y-coordinate, 1).

Absolute value bars ensure the area is always positive, as determinants can be negative depending on vertex order.

Option 2 is missing the $\frac{1}{2}$ factor, which gives double the actual area.

Option 3 has 0 instead of 1 in the last row. The third column must be all ones.

Option 4 has two zeros in the last column, which breaks the formula structure completely.

The correct answer is Option 1.