Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\\left(1-m\right)x+m+2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\\left(1-m\right)x=-m-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\x=\dfrac{m+2}{m-1}\end{matrix}\right.\)
=>\(A\left(\dfrac{m+2}{m-1};0\right)\)
\(OA=\sqrt{\left(\dfrac{m+2}{m-1}-0\right)^2+\left(0-0\right)^2}=\left|\dfrac{m+2}{m-1}\right|\)
Tọa độ B là:
\(\left\{{}\begin{matrix}x=0\\y=0\left(1-m\right)+m+2=m+2\end{matrix}\right.\)
=>B(0;m+2)
\(OB=\sqrt{\left(0-0\right)^2+\left(m+2-0\right)^2}=\left|m+2\right|\)
Để ΔAOB vuông cân thì OA=OB
=>\(\left|m+2\right|=\dfrac{\left|m+2\right|}{\left|m-1\right|}\)
=>\(\left|m+2\right|\left(\dfrac{1}{\left|m-1\right|}-1\right)=0\)
=>\(\left[{}\begin{matrix}m=-2\\\left|m-1\right|=1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=-2\\m-1=1\\m-1=-1\end{matrix}\right.\)
=>\(m\in\left\{-2;0;2\right\}\)
