Let$~\mathbb{R}^2$ be equipped with the usual Euclidean metric, define a line to be a subset of $\mathbb{R}^2$ which is isometric to $\mathbb{R}$. Is it true that all lines must be of the form $y=ax+b$ or be vertical(the more familiar definition)?
By definition a subset l of $\mathbb{R}^2$ (equipped with the Euclidean metric) is said to be isometric to $\mathbb{R}$ iff there is a bijection $f$ from $\mathbb{R}\to l$ such that for all $x,y\in \mathbb{R}$ , $|x-y|=d_E(f(x),f(y))$ (in other words there exists a bijection is which distance preserving). Here $d_{E}$ is the Euclidean metric.
I think the answer to the question is yes, I originally wanted to prove that a line(with the isometry definition) is uniquely determined by two points and I figured that this immediately follows if the answer to my question is "Yes".The proof(if this is true!) must use the Euclidean metric in some fundamental way because the theorem is not generally true, for instance it fails for the taxicab metric.
Best Answer
tl; dr Yes, this is true. Intuitively, if $\ell$ is not a line, there exist three non-collinear points on $\ell$, which is incompatible with $\ell$ being the image of an isometry.
$\newcommand{\Reals}{\mathbf{R}}$One (standard) proof is to let $\ell$ be an arbitrary line, and to fix a distance-preserving bijection $f:\Reals \to \ell$. Pick distinct points $p$ and $q$ of $\ell$ arbitrarily, and let $L$ denote the Euclidean line through $p$ and $q$. Finally, let $v$ be the unit vector parallel to the displacement $q - p$, and let $n$ be either of the two unit vectors orthogonal to $v$. (This is where we're using the Euclidean metric and its origins with an inner product.)
It suffices to show the image of $f$ is contained in $L$.
By hypothesis, there exist real numbers $a$ and $b$ such that $p = f(a)$ and $q = f(b)$. By exchanging the names of $p$ and $q$ if necessary, we may assume $a < b$ without loss of generality. The isometry condition on $f$ implies $t_{0} - a = d_{E}(p, f(t_{0}))$ and $b - t_{0} = d_{E}(f(t_{0}), q)$ provided $a \leq t_{0} \leq b$.
Introduce real-valued functions $$ f_{1}(t) = (f(t) - p) \cdot v,\qquad f_{2}(t) = (f(t) - p) \cdot n. $$ Geometrically, we're resolving $f$ into components with the origin at $p$ and the Cartesian axes parallel to $v$ and to $n$, respectively. It suffices to show $f_{2}$ is identically $0$.
If $t_{0}$ is an arbitrary real number in $(a, b)$, then since $b - a = (t_{0} - a) + (b - t_{0})$ and $f$ is an isometry, \begin{align*} d_{E}(p, q) &= d_{E}(p, f(t_{0})) + d_{E}(f(t_{0}), q) \\ &= \bigl[(f_{1}(t_{0}) - a)^{2} + f_{2}(t_{0})^{2}\bigr]^{1/2} + \bigl[(b - f_{1}(t_{0}))^{2} + f_{2}(t_{0})^{2}\bigr]^{1/2} \\ &\geq |f_{1}(t_{0}) - a| + |b - f_{1}(t_{0})| \geq b - a = d_{E}(p, q); \end{align*} consequently, the inequalities are equalities, and the first implies $f_{2}(t_{0}) = 0$. (The second can be used to show $f_{1}(t) = t - a$, so that $f$ is an affine parametrization of $\ell$.)
We conclude that the image of $[a, b]$ is contained in $L$. Since $p$ and $q$ were arbitrary points of $\ell$, we have $\ell \subset L$.
The reverse inclusion is easier: If $p$ is an arbitrary point of $L$, then $d_{E}(f(0), p) = r$ is a real number. If $r = 0$, then $p = f(0) \in \ell$. Otherwise, there are precisely two points of $L$ at Euclidean distance $r$ from $f(0)$, one of them being $p$, and precisely two real numbers of absolute value $r$. Since $f:\Reals \to \ell$ is a bijection, we have either $p = f(r)$ or $p = f(-r)$. In either case, $p \in \ell$.