Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\\left(m-1\right)x+4=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\x\left(m-1\right)=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{m-1}\\y=0\end{matrix}\right.\)
=>\(A\left(-\dfrac{4}{m-1};0\right)\)
Tọa độ B là:
\(\left\{{}\begin{matrix}x=0\\y=\left(m-1\right)\cdot x+4=0\cdot\left(m-1\right)+4=4\end{matrix}\right.\)
=>B(0;4)
O(0;0); B(0;4); \(A\left(-\dfrac{4}{m-1};0\right)\)
\(OA=\sqrt{\left(-\dfrac{4}{m-1}-0\right)^2+\left(0-0\right)^2}=\sqrt{\left(\dfrac{4}{m-1}\right)^2}=\dfrac{4}{\left|m-1\right|}\)
\(OB=\sqrt{\left(0-0\right)^2+\left(4-0\right)^2}=\sqrt{0+16}=4\)
Vì Ox\(\perp\)Oy
nên OA\(\perp\)OB
=>ΔOAB vuông tại O
=>\(S_{AOB}=\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}\cdot4\cdot\dfrac{4}{\left|m-1\right|}=\dfrac{8}{\left|m-1\right|}\)
Để \(S_{OAB}=2\) thì \(\dfrac{8}{\left|m-1\right|}=2\)
=>|m-1|=8/2=4
=>\(\left[{}\begin{matrix}m-1=4\\m-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=5\\m=-3\end{matrix}\right.\)
