a) \(mx-y=2m-1\Leftrightarrow y=mx-2m+1\left(a=m;b=-2m+1\right)\)
Để (d) đi qua góc tọa độ thì: \(b=0\Rightarrow-2m+1=0\Leftrightarrow m=\dfrac{1}{2}\)
b)
\(mx-y=2m-1\\ \Leftrightarrow mx-2m-y+1=0\\ \Leftrightarrow m\left(x-2\right)-\left(y-1\right)=0\\ \Leftrightarrow m\left(x_0-2\right)-\left(y_0-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_0=2\\y_0=1\end{matrix}\right.\)
=> (d) luôn đi qua điểm (2;1) cố định
c) (d) cắt trục tung tại: \(\left(0;-2m+1\right)=>AO=\left|-2m+1\right|\)
(d) cắt trục tung tại: \(\left(\dfrac{2m-1}{m};0\right)=>BO=\left|\dfrac{2m-1}{m}\right|\)
\(S_{AOB}=\dfrac{1}{2}AO\cdot OB=\dfrac{1}{2}\left|-2m+1\right|\cdot\left|\dfrac{2m-1}{m}\right|\)
\(=\left|\dfrac{1}{2}\left(-2m+1\right)\dfrac{2m-1}{m}\right|\\ =\left|\dfrac{1}{2}\cdot\dfrac{-\left(2m-1\right)^2}{m}\right|\\ =\left|\dfrac{-\left(2m-1\right)^2}{2m}\right|=\dfrac{\left(2m-1\right)^2}{\left|2m\right|}\)
Để \(S_{AOB}=4=>\dfrac{\left(2m-1\right)^2}{\left|2m\right|}=4\Leftrightarrow4m^2-4m+1=8\left|m\right|\)
TH1: m≥0 \(\Rightarrow4m^2-12m+1=0\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{3+2\sqrt{2}}{2}\\m=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\left(tm\right)\)
TH2: m<0 \(\Rightarrow4m^2+4m+1=\Leftrightarrow m=-\dfrac{1}{2}\left(tm\right)\)
