Determine the third point in right triangle only knowing the coordinates of the other two points

There are many solutions for $B$:

Draw a circle trough the points $A$ and $C$, with diameter $|AC|$, then all points on that circle except $A$ and $C$ are solutions

This is called Thales' theorem

We can find the points on this circle by first finding the equation of that circle with center

$$O=\dfrac{A+C}{2}$$

Taking your example: $A=(4,3)$ and $C=(2,1)$ we find that

$$O=\dfrac{(4,3)+(2,1)}{2}=\dfrac{(6,4)}{2}=(3,2)$$

And the radius of the circle is $|AO|=\sqrt{1^2+1^2}=\sqrt{2}$

So the equation of that circle is $$(x-3)^2+(y-2)^2=2$$

All points $B=(x,y)$ that satisfy this equation, except $A$ and $C$ make a right triangle with your given points.

Let's solve the equation for $y$:

$$y=\pm \sqrt{2-(x-3)^2}+2$$

So choose a value for $x$ but make sure the part under the square root will not be negative, and this will give you two valid values for $y$!

Example: choose $x=3$ then the formula gives $y=\pm \sqrt{2}+2$ so $$B=(3,\sqrt{2}+2)$$

Is one of many solutions.