Q.Find the area of the triangle whose vertices are $(2,1)$, $(4,5)$ and $(6,3)$, using determinants.

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

Using $\text{Area} = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|$ with $(2,1)$, $(4,5)$, $(6,3)$: $\begin{vmatrix} 2 & 1 & 1 \\ 4 & 5 & 1 \\ 6 & 3 & 1 \end{vmatrix}.$

Expanding along the third column (entries all $1$): $= 1\begin{vmatrix} 4 & 5 \\ 6 & 3 \end{vmatrix} - 1\begin{vmatrix} 2 & 1 \\ 6 & 3 \end{vmatrix} + 1\begin{vmatrix} 2 & 1 \\ 4 & 5 \end{vmatrix} = (12-30) - (6-6) + (10-4) = -18 - 0 + 6 = -12.$

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

The area can equally be found by the standard formula (1/2)|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| shown as the verification - it i …

Forgetting to take the absolute value before halving is a common slip. A determinant can be negative depending on the order the vertices are lis …