Coordinates of the incenter of a triangle

$$A = (x_{1},y_{1})$$, $$B =(x_{2},y_{2})$$, and $$C = (x_{3},y_{3})$$ are three (distinct) non-collinear points in the Cartesian plane, and $$a = \left\vert \overline{BC} \right\vert$$, $$b = \left\vert \overline{AC} \right\vert$$, and $$c = \left\vert \overline{AB} \right\vert$$. The incenter of the triangle is $$\begin{equation*} \left(\frac{ax_{1} + bx_{2} + cx_{3}}{a + b + c} , \ \frac{ay_{1} + by_{2} + cy_{3}}{a + b + c}\right) . \end{equation*}$$

The $$x$$-coordinate of the incenter is a "weighted average" of the $$x$$-coordinates of the vertices of the given triangle, and the $$y$$-coordinate of the incenter is the same "weighted average" of the $$y$$-coordinates of the same vertices. I am requesting an explanation for this statement.

The bisector of angle $$A$$ intersects side $$BC$$ at a point $$A'$$, and according to angle bisector theorem we have: $$A'B:A'C=c:b$$. It follows that $$A'$$ is a weighted average of $$B$$ and $$C$$, with weights given by the lengths of the opposite sides: $$A'={b\over b+c}B+{c\over b+c}C,$$ and of course we have analogous expressions for the similarly defined points $$B'$$ and $$C'$$.
The incenter $$I$$ of $$ABC$$ is the intersection of $$AA'$$, $$BB'$$ and $$CC'$$. It is then hardly surprising that it turns out to be the weighted average of $$A$$, $$B$$ and $$C$$. For instance: as $$I$$ belongs to segment $$AA'$$ we can write: $$I=(1-t)A+tA'=(1-t)A+{tb\over b+c}B+{tc\over b+c}C,$$ for some $$t\in[0,1]$$. But the expression for $$I$$ must be symmetric when exchanging $$A$$, $$B$$, $$C$$ among them, and it is easy to verify that $$t={b+c\over a+b+c}$$ does the trick, leading to your formula for $$I$$: $$I={a\over a+b+c}A+{b\over a+b+c}B+{c\over a+b+c}C.$$
• For example, if $\overline{AA_{1}}$ is not a vertical line, ... – A gal named Desire Apr 17 at 18:27
• ... $m_{A} = \frac{b(y_{2} - y_{1}) + c(y_{3} - y_{1})}{b(x_{2} - x_{1}) + c(x_{3} - x_{1})}$ ... – A gal named Desire Apr 17 at 18:28