Given three arbitrary points in the euclidean plane, is there a way to homeomorphic map the plane so that the points become colinear? Suppose $p,q,r \in\mathbb{R}^2$. Is there a homeomorphism $h:\mathbb{R^2}\rightarrow \mathbb{R}^2$ such that $h(p),h(q),h(r)$ are colinear? I'm interested in whether such an $h$ be found for $n$ points in the plane, but perhaps I can figure that out from the $n=3$ case. 
(Disclaimer: this question is relevant to a question on exam I had in a previous class, which ended weeks ago.)

I think when $n = 3$ the argument can be made quite a bit simpler, so because you want an argument that can be generalized, I'll go straight to $n$.
Pick a coordinate system in the plane so that the $y$-axis is not parallel to any of the lines through two of the points.
Let the points be $(x_1,y_1), \dots, (x_n,y_n)$. Let $f(x)$ be a continuous function whose graph passes through all $n$ points. Then $h \colon (x,y) \mapsto (x,y - f(x))$ is your required homeomorphism.
