b.
(d) cắt Ox, Oy tại 2 điểm phân biệt khi \(\left\{{}\begin{matrix}3m-2\ne0\\m-2\ne0\end{matrix}\right.\) \(\Rightarrow m\ne\left\{\dfrac{2}{3};2\right\}\)
Khi đó: \(y_A=0\Rightarrow\left(3m-2\right)x_A+m-2=0\Rightarrow x_A=\dfrac{2-m}{3m-2}\)
\(\Rightarrow OA=\left|x_A\right|=\left|\dfrac{2-m}{3m-2}\right|=\left|\dfrac{m-2}{3m-2}\right|\)
\(x_B=0\Rightarrow y_B=\left(3m-2\right).0+m-2=m-2\)
\(\Rightarrow OB=\left|y_B\right|=\left|m-2\right|\)
Tam giác OAB vuông tại O nên:
\(S_{OAB}=\dfrac{1}{2}OA.OB=\dfrac{1}{2}.\left|\dfrac{m-2}{3m-2}\right|.\left|m-2\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{\left(m-2\right)^2}{\left|3m-2\right|}=1\Rightarrow\left[{}\begin{matrix}\left(m-2\right)^2=3m-2\\\left(m-2\right)^2=-3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m^2-7m+6=0\\m^2-m+2=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m=1\\m=6\end{matrix}\right.\)
`b.`
`@` Cho \(x=0\Rightarrow y=m-2\)
\(\Rightarrow A\left(0;m-2\right)\in\left(d\right)\)
`@` Cho \(y=0\Rightarrow x=\dfrac{2-m}{3m-2}\)
\(\Rightarrow B\left(\dfrac{2-m}{3m-2};0\right)\in\left(d\right)\)
Vậy \(\left(d\right)\) cắt trục Ox tại \(B\left(\dfrac{2-m}{3m-2};0\right)\)
\(\left(d\right)\) cắt trục Oy tại \(A\left(0;m-2\right)\)
\(\Rightarrow OB=\left|\dfrac{2-m}{3m-2}\right|;OA=\left|m-2\right|\)
Ta có:
\(S_{OAB}=\dfrac{1}{2}.OA.OB\)
\(\Leftrightarrow\dfrac{1}{2}=\dfrac{1}{2}.\left|m-2\right|.\left|\dfrac{2-m}{3m-2}\right|\) ; \(\left(m\ne\dfrac{2}{3}\right)\)
\(\Leftrightarrow\left|\left(m-2\right)\left(\dfrac{2-m}{3m-2}\right)\right|=1\)
\(\Leftrightarrow\left|-\dfrac{\left(m-2\right)^2}{3m-2}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{\left(m-2\right)^2}{3m-2}=1\\-\dfrac{\left(m-2\right)^2}{3m-2}=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m^2-m+2=0\left(vô.lý\right)\\m^2-7m+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=6\\m=1\end{matrix}\right.\) ( tm )
Vậy \(m=\left\{6;1\right\}\) thì \(S_{OAB}=\dfrac{1}{2}\)
