2: Tọa độ A là:
\(\left\{{}\begin{matrix}x=0\\y=\left(m-2\right)x+m-1=0\left(m-2\right)+m-1=m-1\end{matrix}\right.\)
=>A(0;m-1)
Tọa độ B là:
\(\left\{{}\begin{matrix}y=0\\\left(m-2\right)x+m-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x\left(m-2\right)=-\left(m-1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{-\left(m-1\right)}{m-2}\\y=0\end{matrix}\right.\)
Vậy: \(B\left(\dfrac{-m+1}{m-2};0\right)\)
\(OA=\sqrt{\left(0-0\right)^2+\left(m-1-0\right)^2}=\sqrt{0+\left(m-1\right)^2}=\sqrt{\left(m-1\right)^2}=\left|m-1\right|\)
\(OB=\sqrt{\left(\dfrac{-m+1}{m-2}-0\right)^2+\left(0-0\right)^2}\)
\(=\sqrt{\left(-\dfrac{m-1}{m-2}\right)^2+0}=\left|\dfrac{m-1}{m-2}\right|\)
Vì Ox\(\perp\)Oy
nên OA\(\perp\)OB
=>\(S_{OAB}=\dfrac{1}{2}\cdot\left|m-1\right|\cdot\dfrac{\left|m-1\right|}{\left|m-2\right|}=\dfrac{1}{2}\cdot\dfrac{\left(m-1\right)^2}{\left|m-2\right|}\)
Để \(S_{OAB}=1\) thì \(\dfrac{1}{2}\cdot\dfrac{\left(m-1\right)^2}{\left|m-2\right|}=1\)
=>\(\left(m-1\right)^2=2\left|m-2\right|\)(1)
TH1: m>=2
Phương trình (1) sẽ trở thành: \(\left(m-1\right)^2=2\left(m-2\right)\)
=>\(m^2-2m+1-2m+4=0\)
=>\(m^2-4m+5=0\)
=>\(\left(m-2\right)^2+1=0\)(vô lý)
TH2: m<2
Phương trình (1) sẽ trở thành:
\(\left(m-1\right)^2=2\left(-m+2\right)\)
=>\(m^2-2m+1=-2m+4\)
=>m2=3
=>\(\left[{}\begin{matrix}m=\sqrt{3}\left(nhận\right)\\m=-\sqrt{3}\left(nhận\right)\end{matrix}\right.\)
