Example 2 Given the three vertices of $\triangle A B C$ with coordinates $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$. The lengths of the three sides are $a, b, c$. Find the coordinates of the incenter $I$ of this triangle. (Figure 6-1)
In $\triangle A B C$, since $A D$ bisects $\angle B A C$, according to the property of the internal angle bisector, we have
$$
\frac{|B D|}{|D C|}=\frac{|A B|}{|A C|}=\frac{c}{b} \text {. }
$$

Thus, $x_{D}=\frac{x_{2}+\frac{c}{b} x_{3}}{1+\frac{c}{b}}=\frac{b x_{2}+c x_{3}}{b+c}, y_{D}=\frac{b y_{2}+c y_{3}}{b+c}$.
Also, $B D=\frac{a c}{b+c}$.
In $\triangle A B D$, $B I$ bisects $\angle A B D \cdot \frac{|A I|}{|I D|}=\frac{|A B|}{|B D|}=\frac{b+c}{a}$.
The ratio $\lambda=\frac{b+c}{a}$ that $I$ divides $\overline{A D}$. Using the section formula again, we have
$$
x_{1}=\frac{x_{1}+\frac{b+c}{a} \cdot x_{D}}{1+\frac{b+c}{a}}=\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c},
$$

Similarly, $y_{I}=\frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}$.
Thus, the incenter $I$ is $\left(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}\right)$.