Q.Find the area of the triangle whose vertices are $(1,2)$, $(3,4)$ and $(5,0)$, using determinants.

The area of a triangle from its vertices' coordinates is half the absolute value of a $3\times3$ determinant built from those coordinates and a column of $1$s.

Using the formula $\text{Area} = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|$ with $(x_1,y_1)=(1,2)$, $(x_2,y_2)=(3,4)$, $(x_3,y_3)=(5,0)$: $\begin{vmatrix} 1 & 2 & 1 \\ 3 & 4 & 1 \\ 5 & 0 & 1 \end{vmatrix}.$

Expanding along the third column (entries all $1$): $=1\begin{vmatrix}3&4\\5&0\end{vmatrix} - 1\begin{vmatrix}1&2\\5&0\end{vmatrix} + 1\begin{vmatrix}1&2\\3&4\end{vmatrix} = 1(0-20) - 1(0-10) + 1(4-6) = -20+10-2 = -12.$

So $\text{Area} = \frac{1}{2}|-12| = \frac{1}{2}(12) = 6 \text{ square units.}$ …

The area can equally be found using the standard coordinate-geometry formula $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ shown above as the verificati …

Forgetting to take the absolute value before halving is a common slip — since a determinant can come out negative depending on the order the vertices are listed in, a negative area h …