It is a simpler problem, as uses both compass & ruler; but still need help.
Need draw perpendicular bisector of a line segment with the help of desmos, as here, as if with compass with a slightly greater radius (taken $1.3$ times the radius, as by value of parameter $a=1.3$) than the half of length of the line segment.
The issue is what dimension of $y$ to take in #$7$ function to enable full length of the segment between the two intersection points of the two circles, for all possible values of $(h_1, k_1), (h_2, k_2)$.
Also, I hope that taking ruler is logically same as being able to put a formula for calculating $y$ in function #$7$. Please correct me in this part, if wrong.
Edit -- The below ones are the additional & linked questions that arose in the same context:
There are other errors also, that concern with extra segment of arc appearing elsewhere, and hence logic is wrong & needs correction. Also, wrong logic is showing up for some negative values of $h_1, h_2, k_1, k_2$. Although for the few values used by me, obtained by changing slider for $h_1$ in the given graph always gives the bisector, but the issue is the two arcs, and line segment joining them in function #$7$.
Want to find the formula for generating label for the two intersection points.
Guide:
Suppose your two coordinates are $(h_1, k_1)$ for $A$ and $(h_2, k_2)$ for $B$. Let the midpoint is $(h_3,k_3)$ and the gradient of the line connecting $(h_1,k_1)$ and $(h_2, k_2)$ be $-\frac1m$ where $m\ne 0$ (handle the case when $m=0$ separately).
Then the slope of the perpendicular bisector would be $m$.
Let the distance from $(h_1, k_1)$ to $(h_3, k_3)$ be $d$ and the radius of the circles be $r$, then by Pythagoras theorem, the distance from the intersection point to the midpoint would be $\sqrt{r^2-d^2}.$
Hence the coordinate of the intersection point would be
$$\left(h_3+\frac{\sqrt{r^2-d^2}}{\sqrt{1+m^2}}, k_3+\frac{m\sqrt{r^2-d^2}}{\sqrt{1+m^2}}\right)$$ and
$$\left(h_3-\frac{\sqrt{r^2-d^2}}{\sqrt{1+m^2}}, k_3-\frac{m\sqrt{r^2-d^2}}{\sqrt{1+m^2}}\right).$$
As for your second question, function $7$ is mentioned, are you implying that you are working with a fixed coordinate system, and you can read off any coordinate of any point any connect lines between them. In that case, writing out the equation would of course construct a line and given a line, we can write its equation as well. However, in the classical setting, I don't think we have the abiity to read off the coordinate system. We do not have a ruler, what we have is a straightedge.