On existence of general points in the plane using a device I'm interested in the following problem:

Jupiter is a device that, when given two distinct points $U$ and $V$ in the plane, Jupiter draws the perpendicular bisector of $UV$. If three lines forming a triangle are drawn, can Jupiter mark any point in the plane of the triangle using this device, a pencil, and no other tools? If a point can be marked using the device and a pencil, we call it a Jupiter Point. So, how to determine if a point is Jupiter point or not?

Its pretty easy to see that we can plot the circumcenter of the triangle by simply considering the intersection of perpendicular bisectors of the sides of the triangle. This further gives us the mid-points of sides of the triangle too! Further, we can also construct the circumcenter of the medial triangle using Jupiter and thus, we can plot the nine-point center of triangle too! From ELMO 2020 P3, we can also plot the orthocenter of the triangle. However, I wonder if it is possible to locate all the points in the plane.
From above, its clear that $H, N_9, M_{AB},M_{BC},M_{CA}$ are Jupiter points. However, I'm not very sure how to work out for any general point in the plane.
EDIT: Well, we can't mark all points in the plane as pointed out by @lulu, as the plane is uncountable. However, I'm interested in an algorithm that could tell us if for a given point in the plane, it is possible to mark it using the device or not.
Any help will be highly appreciated!
 A: Not a complete answer, but a pretty good start
Preview: For at least one triangle, not every point is a Jupiter point, by elementary arguments. And we provide a test that conclusively shows that a point is not a Jupiter point for this triangle (but in the event the test fails, we do not show that the point is a Jupiter point).
Consider the special case where $A$ is the origin and $B = (1,0)$ and $C = (0, 1)$.
Call a rational number $n/k$ in lowest terms "2-rational" if $k$ is a power of $2$. (And in case you were wondering, $0$ is 2-rational). A point is 2-rational if its coordinates are both 2-rational numbers. A line is 2-rational if it contains two distinct 2-rational points.
All "marked" points (at this point just $A,B,C$ are 2-rational. And all lines are evidently 2-rational as well.
Lemma: (proof left to the reader) sums and products of 2-rational numbers are again 2-rational.
Small theorem: if $\ell$ is a 2-rational line, then it can be expressed as the zero-set of an equation
$$
ax + by + c = 0
$$
where $a, b,c$ are all 2-rational.
Proof: We know $\ell$ contains distinct 2-rational points $A = (p, q)$ and $B = (r, s).$ Picking $a = (q-s), b = (r - p), c = ps - qr$, we see that $A$ satisfies $ax + by + c = 0$, for
\begin{align}
ax + by + c 
&= (q-s)p + (r-p) q + ps - qr \\
&= qp-sp + rq-pq + ps - qr \\
&= -sp + rq + ps - qr \\
&= 0
\end{align}
and similarly for $B$.
Thus all points and lines in the initial drawing are 2-rational, and the lines have 2-rational line-coefficients.
If $\ell$ and $m$ are distinct, non-parallel lines, and both are 2-rational, then their intersection point $C$ is a 2-rational point. The proof is similar to the proof of the small theorem.
Hence at any level of construction, all lines and all marked points are 2-rational.
Thus in at least the case of this triangle, the set of constructable points is quite tiny compared to the set of all points of the plane.
Is every 2-rational point constructible from this triangle? I suspect so, but I don't have the will or energy to prove it. Some construction that lets one carry out the euclidean algorithm in some form (to construct any possible numerator) is all that's really needed.
So the promised "test" for a point $A = (r, s)$ is this:

*

*If either $r$ or $s$ is irrational, then $A$ is not a Jupiter point.


*Express each of $r$ and $s$ as a fraction in lowest terms. If either denominator is not an integer power of $2$, then $A$ is not a Jupiter point.


*If both steps 1 and 2 fail, then no conclusion can be drawn (yet).
