It is well known that every planar graph has an embedding such that every edge is drawn as a straight line segment (Fáry's Theorem).  Kemnitz and Harborth made the following stronger conjecture

**Conjecture 1.** Every planar graph has a straight line embedding with *integer* edge lengths.  

I was wondering if it is possible to attack this problem with the following approach.

**Conjecture 2.** Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear.  For any $\epsilon >0$, there exists $X':=\{x_1', \dots, x_n'\}$ such that for all $i, j \in [n]$

1. $d(x_i, x_i') < \epsilon$,
2. $d(x_i', x_j') \in \mathbb{Q}$, and
3. no three points of $X'$ are colinear.

To prove Conjecture 2, it suffices to prove the following conjecture.

**Conjecture 3.** Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear and all pairwise distances are rational.  Then the set of points which are at rational distance from all points in $X$ is a dense subset of the plane.

Conjecture 3 is trivial for $n=1$ and easy for $n=2$.  Almering proved it for $n=3$, and I think it is open for $n>3$.  Note that Conjecture 3 is (essentially) a weakening of:

**Conjecture 4.** There exists a dense subset of the plane with all pairwise distances rational.

This question was posed by Ulam in 1945 (see [this](https://mathoverflow.net/questions/19127/is-there-a-dense-subset-of-the-real-plane-with-all-pairwise-distances-rational/19129#19129) mathoverflow question for more background).  So, the reason I like Conjecture 3 is that it is still strong enough to prove Conjecture 1, but appears much weaker than Conjecture 4.
Unfortunately, Conjecture 3 is beyond my limited area of expertise.  Hence:

>**Question.** What are the prospects for proving Conjecture 3?  A proof or disproof would be fantastic.  However, even arguments suggesting that it is true/false but say beyond current technology would be most welcome.