As the title suggests, I would like to solve for the foot of perpendicular involving complex numbers. In fact, I have already worked out the solutions, but I need help in understanding certain parts of my workings.
Suppose that the circumcircle of triangle A$BC$ is the unit circle. We use the complex numbers $a$, $b$ and $c$ to represent the vertices as mentioned. Also, let W be an arbitrary point on the circle. Clearly, $|a|=|b|=|c|=|w|=1.$
I would like the find the foot of perpendicular, denoted by $P$ from $W$ to the line $AB$. This is how I've done it:
The equation of a line passing through arbitrary points $s_1$ and $s_2$ in complex form is well known and is given by: $$z+s_1s_2\bar{z}=s_1+s_2.$$
In our question, since $p$ lies on the line $AB$, we have that $p$ must satisfy $$p+ab\bar{p}=a+b \text{ -------$(\star)$}$$ and $$p-ab\bar{p}=w-ab\bar{w}\text{ -------$(\star \star)$}$$
Solving both $(\star)$ and $(\star \star)$ yields $$p=\dfrac{a+b+w-ab\bar{w}}{2}.$$
I sought for some help to obtain $(\star \star)$ and I am unable to understand how it is obtained. Is it supposed to be the equation of perpendicular bisector of the line $AB$, passing through the points $W$ and $P$? I am mainly confused by the negative sign as it is not our usual definition of a line. Also, I have tried to expand $|z-z_1|=|z-z_2|$ (eqn of bisector) but it does not match up. Any help would be appreciated!