Question

[Tam Akm]

Cho đường thẳng (\(d_m\))có phương trình(\(d_m\)):y=\(-\dfrac{m-1}{2m-3}\)x +\(\dfrac{m+1}{2m-3}\)
1)xác định m để 
a)(\(d_m\))đi qua A(2;1)
b)(\(d_m\))có hướng đi lên (hàm số đồng biến)
c)(\(d_m\))song song với đường thẳng(△):x-2y=12=0
2)tìm điểm cố định mà (\(d_m\))luôn đi qua

Answer 1

[Nguyễn Hoàng Minh]

\(1,\\ a,A\left(2;1\right)\in\left(d_m\right)\Leftrightarrow\dfrac{-2\left(m-1\right)+m+1}{2m-3}=1\\ \Leftrightarrow-2m+2+m+1=2m-3\\ \Leftrightarrow3m=6\Leftrightarrow m=2\\ b,\Leftrightarrow-\dfrac{m-1}{2m-3}>0\Leftrightarrow\dfrac{m-1}{2m-3}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m-1>0\\2m-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}m-1< 0\\2m-3>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< m< \dfrac{3}{2}\\ c,\left(\Delta\right):x-2y-12=0\Leftrightarrow2y=x-12\Leftrightarrow y=\dfrac{1}{2}x-6\\ \left(d_m\right)\text{//}\left(\Delta\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1-m}{2m-3}=\dfrac{1}{2}\\\dfrac{m+1}{2m-3}\ne-6\end{matrix}\right.\Leftrightarrow m=\dfrac{5}{4}\)

\(2,\text{Gọi }M\left(x_0;y_0\right)\text{ là điểm cần tìm}\\ \Leftrightarrow y_0=\dfrac{1-m}{2m-3}x_0+\dfrac{m+1}{2m-3}\\ \Leftrightarrow y_0\left(2m-3\right)=x_0\left(1-m\right)+m+1\\ \Leftrightarrow x_0-mx_0+m+1-2my_0-3y_0=0\\ \Leftrightarrow m\left(1-x_0-2y_0\right)+\left(x_0-3y_0+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x_0+2y_0=1\\x_0-3y_0=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0=\dfrac{1}{5}\\y_0=\dfrac{2}{5}\end{matrix}\right.\\ \Leftrightarrow M\left(\dfrac{1}{5};\dfrac{2}{5}\right)\)