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enter image description hereAs the title suggests, I would like the find the equation of the perpendicular line from $M_1$ of the side $A_2A_3$ to the tangent line at $A_1$. Without a loss of generality, suppose that on the unit circle, the coordinates of $A_1$, $A_2$ and $A_3$ are given by $u_1$, $u_2$ and $u_3$ respectively.

It is well known that the complex form of a line passing through 2 distinct point $p_1$ and $p_2$ is given by $$z+ p_1p_2\bar{z}=p_1+p_2.$$
Since the tangent to the unit circle at $A_1$ coincides, it is clear that the equation of tangent at $A_1$ is given by $$z+u_1^2\bar{z}=2u_1.$$ How can I find the equation of the perpendicular line from $M_1$(where $M_1=\dfrac{u_2+u_3}{2})$ of the side $A_2A_3$ to the tangent line at $A_1$?

When I was reading the proof, the author gave the required perpendicular line to be $$z-u_1^2\bar{z}=\frac{1}{2}[(u_2+u_3)-u_1^2(\bar{u_2}+\bar{u_3})].$$

Can someone explain to me how is this obtain? I am mainly confused as to how the sign between $z$ and $\bar{z}$ changed from positive to negative. Also, the author seem to ignore the "constant" terms on the RHS (That is, the "$p_1+p_2$" constant for line equations) and simply subbed in $M_1=\dfrac{u_2+u_3}{2}$ into the LHS. Any help would be appreciated!

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The equation of a perpendicular line from $C(c)$ to a line $AB$ where $A(a),B(b)$ with $|a|=|b|=1$ and $a\not =b$ is given by $$\begin{align}\frac{z-c}{a-b}\in\mathbb Ri &\iff (z-c)(\bar a-\bar b)\in\mathbb Ri \\\\&\iff (z-c)\left(\frac 1a-\frac 1b\right)+(\bar z-\bar c)(a-b)=0 \\\\&\iff z-ab\bar z=c-ab\bar c\end{align}$$

Now, set $a=u_1,c=\frac{u_2+u_3}{2}$ and take $b\to a$.

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  • $\begingroup$ thank you so much, you're a saviour. Have a great day! $\endgroup$
    – Cleytus
    Commented Sep 22, 2018 at 7:59

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