Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\\left(6-4m\right)x+3-m=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=\dfrac{m-3}{6-4m}\end{matrix}\right.\)
=>OA=|m-3|/|4m-6|
Tọa độ B là:
\(\left\{{}\begin{matrix}x=0\\y=\left(6-4m\right)\cdot0+3-m=3-m\end{matrix}\right.\Leftrightarrow OB=\left|m-3\right|\)
Theo đề, ta có: \(OA=OB\)
=>|m-3|/|4m-6|=|m-3|
\(\Leftrightarrow\left(\left|m-3\right|\right)\cdot\left(\dfrac{1}{\left|4m-6\right|}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m-3=0\\\left|4m-6\right|=1\end{matrix}\right.\Leftrightarrow m\in\left\{3;\dfrac{7}{4};\dfrac{5}{4}\right\}\)