Prove that a metric space with exactly 3 points is isometric to a subset of E^2.

Question

I do not know how to proof this. The only thing I can think of is that $E^2$ is isomatric with $R^2$. And that because of metric space X you know there's a triangle inequality with $x,y,z\in X$. With $E^2$ the Euclidean space and $R^2$ the two dimensional real vectors.

Answer 1

Hint: without loss of generality, you can arrange for your isometry to place $x$ at $(0, 0)$ and $y$ at $(0, d(y-x))$. Now use the triangle inequality to show that you have two equally good choices for where to place $z$.