b: Tọa độ điểm A là:
\(\left\{{}\begin{matrix}x=0\\y=3\cdot0+2=2\end{matrix}\right.\Leftrightarrow A\left(0;2\right)\)
Tọa độ điểm B là:
\(\left\{{}\begin{matrix}y=0\\3x+2=0\end{matrix}\right.\Leftrightarrow B\left(-\dfrac{2}{3};0\right)\)
\(OA=\sqrt{\left(0-0\right)^2+\left(2-0\right)^2}=2\)
\(OB=\sqrt{\left(-\dfrac{2}{3}-0\right)^2}=\dfrac{2}{3}\)
\(S_{OAB}=\dfrac{OA\cdot OB}{2}=\dfrac{4}{3}:2=\dfrac{4}{6}=\dfrac{2}{3}\)