Find cartesian coordinates of the incenter $A(a_1,a_2)$, $B(b_1,b_2)$ and $C(c_1,c_2)$ form the triangle $ABC$. What are the cartesian coordinates of the incenter and why?
 A: I'll argue up to the formula, so the "why" will be clear. Let's use this diagram of $\triangle ABC$ with the Cartesian coordinates moved so that points $A$ and $B$ are on the $x$-axis and $C$ has a positive $y$-coordinate.

Point $I$ is the incenter, segments $ID$, $IE$, and $IF$ are perpendiculars from $I$ to sides $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ respectively. Since $I$ is the incenter, those perpendicular segments are radii of the incircle and have equal lengths, which we call $r$. The area of $\triangle ABC$ is of course half-times-base-times-height, with the base here $c$ and height $c_2$. We then get
$$\begin{align}
\operatorname{Area}(\triangle ABC)
 &= \operatorname{Area}(\triangle ABI)+\operatorname{Area}(\triangle BCI)
   +\operatorname{Area}(\triangle ACI) \\[2 ex]
 &= \frac 12cr+\frac 12ar+\frac 12br \\[2 ex]
 &= \frac{a+b+c}2r
\end{align}$$
Thus $r$, which is the $y$-coordinate of point $I$, is given by
$$\begin{align}
r&=\frac{2\operatorname{Area}(\triangle ABC)}{a+b+c} \\[2 ex]
 &=\frac{2\cdot\frac 12c\cdot c_2}{a+b+c} \\[2 ex]
 &=\frac{c\cdot c_2}{a+b+c}
\end{align}$$
If we look only at the $y$-coordinates, and consider $A$, $B$, $C$, and $I$ to be vectors, we get

$$I=\frac{aA+bB+cC}{a+b+c}$$

Note that this formula does not use the coordinate system directly and is symmetrically dependent on the triangle's vertices. Through an argument too tedious to give here, we get that the formula for $I$ is true in both $x$- and $y$-coordinates in any Cartesian coordinate system.
If you really want a formula with only the coordinates $a_1,a_2,b_1,b_2,c_1,c_2$, we replace $A$ with $(a_1,a_2)$ and $a$ with $\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}$ etc., and separate the $x$- and $y$-coordinates, to get

$$I=$$
$$\left(
\frac{a_1\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}
+b_1\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}
+c_1\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}
{\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}
+\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}
+\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}
,\right.$$
$$\left.
\frac{a_2\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}
+b_2\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}
+c_2\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}
{\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}
+\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}
+\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}
\right)$$

