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

The area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is the absolute value of $\tfrac12\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}$. With $(0,0),(1,2),(4,3)$ the determinant expands to $-5$, so the area is $\tfrac12|-5|=\tfrac52$.

The area of a triangle is a classic application of a $3\times3$ determinant: for vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ it equals the absolute value of $\tfrac12\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}$.

Step 1. Substitute the vertices $(0,0),(1,2),(4,3)$:

Step 2. Expand along the first row (it has two zeros, so only the third entry contributes):

Step 3. Take half the absolute value:

Area-as-determinant formula (absolute value of half the 3×3 vertex determinant).

• Forgetting to take the absolute value — a determinant can be negative, but area is always non-negative.
• Dropping the factor of $\tfrac12$ in front of the determinant.
• Misplacing a vertex's coordinates into the wrong determinant column.