Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\\left(m+1\right)x-2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\x\left(m+1\right)=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=\dfrac{2}{m+1}\end{matrix}\right.\)
=>\(A\left(\dfrac{2}{m+1};0\right)\)
\(OA=\sqrt{\left(\dfrac{2}{m+1}-0\right)^2+\left(0-0\right)^2}=\sqrt{\left(\dfrac{2}{m+1}\right)^2}=\dfrac{2}{\left|m+1\right|}\)
Tọa độ B là:
\(\left\{{}\begin{matrix}x=0\\y=\left(m+1\right)\cdot x-2=0\cdot x-2=-2\end{matrix}\right.\)
=>B(0;-2)
\(OB=\sqrt{\left(0-0\right)^2+\left(-2-0\right)^2}=\sqrt{0^2+\left(-2\right)^2}=2\)
Ox\(\perp\)Oy
=>OA\(\perp\)OB
=>ΔOAB vuông tại O
Để góc OAB=45 độ thì ΔOAB vuông cân tại O
=>OA=OB
=>\(\dfrac{2}{\left|m+1\right|}=2\)
=>\(\left|m+1\right|=1\)
=>\(\left[{}\begin{matrix}m+1=1\\m+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-2\end{matrix}\right.\)
