Tọa độ A là:
\(\left\{{}\begin{matrix}x+1=-2x\\y=-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=-1\\y=-2x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=-2\cdot\dfrac{-1}{3}=\dfrac{2}{3}\end{matrix}\right.\)
Tọa độ B là:
\(\left\{{}\begin{matrix}y=0\\x+1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)
Vậy: O(0;0); \(A\left(-\dfrac{1}{3};\dfrac{2}{3}\right)\); B(-1;0)
\(OA=\sqrt{\left(-\dfrac{1}{3}-0\right)^2+\left(\dfrac{2}{3}-0\right)^2}=\dfrac{\sqrt{5}}{3}\)
\(OB=\sqrt{\left(-1-0\right)^2+\left(0-0\right)^2}=1\)
\(AB=\sqrt{\left(-1+\dfrac{1}{3}\right)^2+\left(0-\dfrac{2}{3}\right)^2}=\sqrt{\dfrac{4}{9}+\dfrac{4}{9}}=\dfrac{2\sqrt{2}}{3}\)
Xét ΔOAB có
\(cosAOB=\dfrac{OA^2+OB^2-AB^2}{2\cdot OA\cdot OB}=\dfrac{\sqrt{5}}{5}\)
=>\(sinAOB=\sqrt{1-\left(\dfrac{\sqrt{5}}{5}\right)^2}=\dfrac{2}{\sqrt{5}}\)
Diện tích tam giác AOB là:
\(S_{AOB}=\dfrac{1}{2}\cdot OA\cdot OB\cdot sinAOB\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{3}\cdot1=\dfrac{1}{3}\)
