## Thomas' Calculus 13th Edition

$$\dfrac{3}{2}$$
The area of the given triangle is: $Area= \dfrac{1}{2}|\vec{AB}\times\vec{AC}|$ $$\vec{AB}= -\hat{i}+2\hat{j} \\ \vec{AC}= -\hat{i}-\hat{j}$$ Now, $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&2&0\\-1&0&-1\end{vmatrix} \\=-2\hat{i}+\hat{j}+2\hat{k}$ and $|\vec{AB}\times\vec{AC}|=\sqrt {4+1+4}= 3$ So, Area =$\dfrac{3}{2}$