Incenter of a triangle formed by three lines.

Question

How can we find the incenter of a triangle (without using its vertices) that is formed by three lines $y=m_1x + c_1, y=m_2x + c_2, y=m_3x + c_3$?

Answer 1

As in the solution referred to in the comment by @conditionalMethod, the incentre $(u.v)$ must be equidistant from the three lines.

The distance, $d$ say, then satisfies

$$d=\frac{|m_iu-v+c_i|}{\sqrt{m_i^2+1}}.$$

We can express these equations neatly as

$$\begin{pmatrix}-m_1&1&\sqrt{m_1^2+1} \\-m_2&1&\pm \sqrt{m_2^2+1}\\-m_3&1&\pm \sqrt{m_3^2+1}\\\end{pmatrix}\begin{pmatrix}u \\v\\*\\\end{pmatrix}=\begin{pmatrix}c_1 \\c_2\\c_3\\\end{pmatrix}$$

Given values for the $m_i,c_i$ these equations can be solved for $u,v$ in a mechanical manner. A drawback of this method is that the incircle is obtained for only one choice of the +/-. Looking positively at this though, you get the bonus of the equations of the 3 excircles from the other choices.