\(a,m=-\dfrac{1}{3}\Leftrightarrow y=\left(1-\dfrac{2}{3}\right)x-2=\dfrac{1}{3}x-2\\ b,\Leftrightarrow4\left(2m+1\right)=-1\Leftrightarrow2m+1=-4\Leftrightarrow m=-\dfrac{5}{2}\\ c,\text{Gọi điểm đó là }A\left(x_0;y_0\right)\\ \Leftrightarrow y_0=2mx_0+x_0-2\\ \Leftrightarrow2mx_0+x_0-y_0-2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x_0=0\\x_0-y_0-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0=0\\y_0=-2\end{matrix}\right.\Leftrightarrow A\left(0;-2\right)\\ \text{Vậy }A\left(0;-2\right)\text{ là điểm cần tìm}\)
\(b,\text{PT giao Ox và Oy: }\left\{{}\begin{matrix}y=0\Rightarrow x=\dfrac{2}{2m+1}\Rightarrow A\left(\dfrac{2}{2m+1};0\right)\Rightarrow OA=\dfrac{2}{\left|2m+1\right|}\\x=0\Rightarrow y=-2\Rightarrow B\left(0;-2\right)\Rightarrow OB=2\end{matrix}\right.\\ \Delta OAB\text{ vuông cân tại }O\\ \Leftrightarrow\dfrac{2}{\left|2m+1\right|}=2\Leftrightarrow\left|2m+1\right|=1\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-1\end{matrix}\right.\)
