Q.Using determinants, find the area of the triangle whose vertices are $(3,8)$, $(-4,2)$ and $(5,-1)$.

[!TLDR] Use Area $=\tfrac12\left|\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\right|$ with the three g …

With $(x_1,y_1)=(3,8),\ (x_2,y_2)=(-4,2),\ (x_3,y_3)=(5,-1)$: Area $=\tfrac12\begin{vmatrix}3&8&1\\-4&2&1\\5&-1&1\end{vmatrix}$. Expanding along the third column: $=\tfrac12\Big[1\begin{vmatrix}-4&2\\5&-1\end{vmatrix}-1\begin{vmatrix}3&8\\5&-1\end{vmatrix}+1\begin{vmatrix}3&8\\-4&2\end{vmatrix}\Big]$. Now $\begin{vmatrix}-4&2\\5&-1\end{vmatrix}=4-10=-6$; $\begin{vmatrix}3&8\5&-1\end{vmatrix}=-3-40= …

Substitute the three vertices into the determinant area formula, expand the $3\times3$ det …

A frequent error is omitting the final absolute value or the factor of $\tfrac12$, or making a sign slip in one of the thr …