$A = (x_{1},y_{1})$, $B =(x_{2},y_{2})$, 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.
Best Answer
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 $I$: $$ I={a\over a+b+c}A+{b\over a+b+c}B+{c\over a+b+c}C. $$