1: Tọa độ A là:
\(\left\{{}\begin{matrix}2x+4=-x+1\\y=-x+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=-3\\y=-x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\left(-1\right)+1=2\end{matrix}\right.\)
vậy: A(-1;2)
2: O(0;0); A(-1;2); B(-1;-4)
\(OA=\sqrt{\left(-1-0\right)^2+\left(2-0\right)^2}=\sqrt{5}\)
\(OB=\sqrt{\left(-1-0\right)^2+\left(-4-0\right)^2}=\sqrt{17}\)
\(AB=\sqrt{\left(-1+1\right)^2+\left(-4-2\right)^2}=6\)
Xét ΔOAB có \(cosAOB=\dfrac{OA^2+OB^2-AB^2}{2\cdot OA\cdot OB}=\dfrac{5+17-36}{2\cdot\sqrt{5}\cdot\sqrt{17}}=-\dfrac{7}{\sqrt{85}}\)
=>\(sinAOB=\sqrt{1-\left(-\dfrac{7}{\sqrt{85}}\right)^2}=\dfrac{6}{\sqrt{85}}\)
Diện tích tam giác AOB là:
\(S_{AOB}=\dfrac{1}{2}\cdot OA\cdot OB\cdot sinAOB\)
\(=\dfrac{1}{2}\cdot\dfrac{6}{\sqrt{85}}\cdot\sqrt{5}\cdot\sqrt{17}=3\)
