I have this problem and solution but I arrived at a different solution. Where am I wrong? The problem is to compute the area of the next figure:
However, I solved in a different way.
$$\overrightarrow{OA}=\langle1,2\rangle\\ \overrightarrow{OC}=\langle3,-1\rangle\\ \overrightarrow{BA}=\langle-3,-1\rangle\\ \overrightarrow{BC}=\langle-1,-4\rangle\\ \text{Area } \triangle AOC=\frac{1}{2} \begin{vmatrix} 1 & 2 \\ 3 & -1 \\ \end{vmatrix}\\ \text{Area } \triangle ABC=\frac{1}{2} \begin{vmatrix} -3 & -1 \\ -1 & -4 \\ \end{vmatrix}\\ \text{Area } \triangle AOC +\text{Area } \triangle ABC=\frac{-1-6+12-1}{2}=\frac{4}{2}$$
Where am I wrong?
You should see that your areas have different signs, so you subtract one from the other . So either take the absolute value or the or change the order of multiplication of your vectors .