Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\\left(m^2-m+2\right)x+2m-8=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\x=\dfrac{-2m+8}{m^2-m+2}\end{matrix}\right.\)
=>\(OA=\sqrt{\left(\dfrac{-2m+8}{m^2-m+2}-0\right)^2+\left(0-0\right)^2}=\dfrac{\left|2m-8\right|}{m^2-m+2}\)
Tọa độ B là:
\(\left\{{}\begin{matrix}x=0\\y=0\left(m^2-m+2\right)+2m-8=2m-8\end{matrix}\right.\)
Vậy:B(0;2m-8)
\(OB=\sqrt{\left(0-0\right)^2+\left(2m-8-0\right)^2}=\left|2m-8\right|\)
ΔOAB vuông tại O
=>\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB\)
\(=\dfrac{1}{2}\left|2m-8\right|\cdot\dfrac{\left|2m-8\right|}{m^2-m+2}\)
\(=\dfrac{1}{2}\cdot\dfrac{\left(2m-8\right)^2}{m^2-m+2}=\dfrac{1}{2}\cdot\dfrac{4\left(m-2\right)^2}{m^2-m+2}=\dfrac{2\left(m-2\right)^2}{m^2-m+2}\)
\(S_{OAB}=2\)
=>\(\dfrac{2\left(m-2\right)^2}{m^2-m+2}=2\)
=>\(\left(m-2\right)^2=m^2-m+2\)
=>\(m^2-4m+4=m^2-m+2\)
=>-3m=-2
=>\(m=\dfrac{2}{3}\)
