Gọi A,B lần lượt là giao điểm của (d) với trục Ox và Oy
Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\\left(2m+1\right)x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x\left(2m+1\right)=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\x=\dfrac{2}{2m+1}\end{matrix}\right.\)
=>\(A\left(\dfrac{2}{2m+1};0\right)\)
\(OA=\sqrt{\left(\dfrac{2}{2m+1}-0\right)^2+\left(0-0\right)^2}=\sqrt{\left(\dfrac{2}{2m+1}\right)^2}=\dfrac{2}{\left|2m+1\right|}\)
Tọa độ B là:
\(\left\{{}\begin{matrix}x=0\\y=\left(2m+1\right)x-2=0\cdot\left(2m+1\right)-2=-2\end{matrix}\right.\)
=>B(0;-2)
\(OB=\sqrt{\left(0-0\right)^2+\left(-2-0\right)^2}=\sqrt{0+4}=2\)
Vì Ox\(\perp\)Oy
nên OA\(\perp\)OB
=>ΔOAB vuông tại O
=>\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}\cdot2\cdot\dfrac{2}{\left|2m+1\right|}=\dfrac{2}{\left|2m+1\right|}\)
Để \(S_{OAB}=1\) thì \(\dfrac{2}{\left|2m+1\right|}=1\)
=>|2m+1|=2
=>\(\left[{}\begin{matrix}2m+1=2\\2m+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2m=1\\2m=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=\dfrac{1}{2}\\m=-\dfrac{3}{2}\end{matrix}\right.\)