Find an equation of each line normal to the graph y=2x/(x-1) and parallel to the line 2x-y+1=0
$\displaystyle y = \frac{2x}{x - 1} = 2 + \frac{2}{x-1} \Rightarrow \frac{dy}{dx} = \frac{-2}{(x-1)^2}$.
You want the normal to be parallel to $\displaystyle 2x - y + 1 = 0 \Rightarrow y = 2x + 1$.
To find the x-coordinates of the required normal solve $\displaystyle \frac{(x-1)^2}{2} = 2$: x = 0, 2.
Therefore the required normal is normal to the curve at (0, 0) and (2, 4). The equation of this normal is y = 2x.
Note: The normal is also the line of symmetry of the curve.