How many ways to find the center of an inscribed circle?

Question

I want to find the coordinates of center of the inscribed circle triangle $ABC$, where $A(-274,-253)$, $B(-1,7)$, $C(14,7)$. I tried

First way. We have $c = AB=377$, $a = BC=15$, $b = AC=388$. Let $I(x,y)$ be the center, using the formulas $$a \overrightarrow{IA} + b\overrightarrow{IB}+c\overrightarrow{IC}=\overrightarrow{0}$$ we have the system of equations $$\begin{cases} 15(-274-x)+388(-1-x)+377(14-x)=0\\ 15(-253-y)+388(7-y)+377(7-y)=0 \end{cases}.$$ Solve this system of equations, we get $I(1,2)$.

Second way.

1. The equation of the line $AB$ is $260x-273y+2171 = 0$.
2. The equation of the line $AB$ is $260x-288y-1624 = 0$.
3. The equation of the line $BC$ is $y = 7$.
4. The equation of the bisector of the interior at the vertex $A$ is $255x-275y+295 = 0$.
5. The equation of the bisector of the interior at the vertex $B$ is $5x+2y-9 = 0$.
6. The center $I$ is the intersection point of two bisectors, there for we solve the system $$\begin{cases} 255x-275y+295 = 0\\ 5x+2y-9 = 0 \end{cases}.$$ and get $I(1,2)$.

Third way. Using the formulas $$(x_I,y_I)=\left(\frac{ax_a+bx_b+cx_c}{a+b+c},\frac{ay_a+by_b+cy_c}{a+b+c}\right)$$

Are there another ways?

Answer 1

I don't know how many ways there are exactly. But I can tell you another solution.

Letting $|F|$ be the area of a figure $F$, we have

$$|ABC|=|IAB|+|IBC|+|ICA|=\frac 12\times r\times (AB+BC+CA)$$ where $r$ is the radious of the inscribed circle.

Since we can get $|ABC|, AB+BC+CA$, we get $r$. So we now know that the distance between $I$ and the line $AB$ is $r$, and that the distance between $I$ and the line $BC$ is also $r$. Then, calculation will tell you the coordinate of $I$.

As you may notice, this solution is not very good. I'm writing this because you seem to be interested in another solution. To be honest, I would not use my solution unless the shape of the given triangle is something special.