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mong bn thông cảm
nha ................
\(a,\Leftrightarrow y=0;x=2\Leftrightarrow2m-2+m-2=0\Leftrightarrow m=\dfrac{4}{3}\)
\(b,\) PT giao Ox: \(\Leftrightarrow\left(m-1\right)x=2-m\Leftrightarrow x=\dfrac{2-m}{m-1}\Leftrightarrow A\left(\dfrac{2-m}{m-1};0\right)\Leftrightarrow OA=\left|\dfrac{2-m}{m-1}\right|\)
PT giao Oy: \(y=m-2\Leftrightarrow B\left(0;m-2\right)\Leftrightarrow OB=\left|m-2\right|\)
\(S_{OAB}=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{2}OA\cdot OB=\dfrac{2}{3}\Leftrightarrow\left|\dfrac{2-m}{m-1}\cdot\left(m-2\right)\right|=\dfrac{4}{3}\\ \Leftrightarrow\left|\dfrac{-\left(m-2\right)^2}{m-1}\right|=\dfrac{4}{3}\Leftrightarrow\left[{}\begin{matrix}\dfrac{-\left(m-2\right)^2}{m-1}=\dfrac{4}{3}\left(1\right)\\\dfrac{-\left(m-2\right)^2}{1-m}=\dfrac{4}{3}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow-3m^2+12m-12=4m-4\\ \Leftrightarrow3m^2-9m+9=0\\ \Leftrightarrow m\in\varnothing\\ \left(2\right)\Leftrightarrow-3m^2+12m-12=4-4m\\ \Leftrightarrow3m^2-16m+16=0\\ \Leftrightarrow\left[{}\begin{matrix}m=4\\m=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}m=4\\m=\dfrac{4}{3}\end{matrix}\right.\) thỏa đề
\(c,\) Gọi \(E\left(x_0;y_0\right)\) là điểm cần tìm
\(\Leftrightarrow\left(m-1\right)x_0+m-2=y_0\\ \Leftrightarrow mx_0+m-x_0-y_0-2=0\\ \Leftrightarrow m\left(x_o+1\right)-\left(x_0+y_0+2\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x_0=-1\\y_0=-2-x_0=-1\end{matrix}\right.\Leftrightarrow E\left(-1;-1\right)\)