I'm confused because I thought that one cannot assume anything is in a fixed position while on a number line. It says the answer is C but I don't understand how they can deduce that from the given information. Twice as far from $x$ as from $y$? there are no markers so how can one know if it's not drawn to scale?
It doesn't matter if the scale isn't specified all you need to know is that there is some distance between $x$ and $y$ and there are other points on the lines with various distances to $x$ and $y$. Just because you can't assume you know the distance between them does not mean that you need to throw out all your intution.
I recommend this. Place your finger on the far left of the number line. Are points in that region twice as far from $x$ as they are from $y$? To my eyes it looks like they are closer to $x$ than $y$ and therefore do not count.
Move your finger to the right till the points start looking like they are closer to $y$ than $x$. This should be between $x$ and $y$ a bit closer to $y$ than $x$. Can you see that even though you aren't sure exactly where it is there has to be a point in that middle region which is twice as far from $x$ as it is from $y$? In other words that there is a point whose distance to $y$ is half its distance to $x$? I think it should be clear there is one point in the middle that does this.
Now keep moving your finger to the right passing through $y$. Notice that while going from the middle region to the right hand region your distance to $y$ goes to zero and then starts getting bigger again. Can you see that we are about to hit another point where $x$ is twice as far from your finger as $y$ is and that there is only one point on the right which does this? This point is the one that is as far from $y$ as $y$ is from $x$.
Hope that helps.
There will be one point in between $x$ and $y$ and another to the right of $y$. For the first case take $x+\frac{2(y-x)}{3}$ as the point and for the second case the point that does it is $y+(y-x)$.
Take $x$ and $y$ to be any two distinct points on the number line.
Assume $y > x$. You want $2\left| z - y \right| = \left|z - x \right|$. Then if $z>y>x$, $z = (2y - x)>y$. If $y>z>x$ then $z = \frac{1}{3} (x +2y) < y$. If we try $z=y$, $z=x$, or $y>x>z$ then we get no consistent solutions.
