To find the coordinates of the centroid and incenter of triangle ABC with vertices A(-36, 7), B(20, 7), and C(0, -8), we will follow these steps:
### Step 1: Calculate the Centroid
The formula for the centroid (G) of a triangle with vertices at points \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) is given by:
\[
G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
\]
Substituting the coordinates of points A, B, and C:
- \(x_1 = -36\), \(y_1 = 7\)
- \(x_2 = 20\), \(y_2 = 7\)
- \(x_3 = 0\), \(y_3 = -8\)
Calculating the x-coordinate of the centroid:
\[
G_x = \frac{-36 + 20 + 0}{3} = \frac{-16}{3}
\]
Calculating the y-coordinate of the centroid:
\[
G_y = \frac{7 + 7 - 8}{3} = \frac{6}{3} = 2
\]
Thus, the coordinates of the centroid are:
\[
G\left( -\frac{16}{3}, 2 \right)
\]
### Step 2: Calculate the Incenter
To find the incenter (I), we first need to calculate the lengths of the sides of the triangle using the distance formula:
\[
AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
\[
BC = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2}
\]
\[
CA = \sqrt{(x_1 - x_3)^2 + (y_1 - y_3)^2}
\]
Calculating the lengths:
1. Length of AB:
\[
AB = \sqrt{(20 - (-36))^2 + (7 - 7)^2} = \sqrt{(20 + 36)^2} = \sqrt{56^2} = 56
\]
2. Length of BC:
\[
BC = \sqrt{(0 - 20)^2 + (-8 - 7)^2} = \sqrt{(-20)^2 + (-15)^2} = \sqrt{400 + 225} = \sqrt{625} = 25
\]
3. Length of CA:
\[
CA = \sqrt{(-36 - 0)^2 + (7 - (-8))^2} = \sqrt{(-36)^2 + (15)^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39
\]
Now, using the lengths of the sides \(a = BC\), \(b = CA\), and \(c = AB\) in the incenter formula:
\[
I\left( \frac{a x_1 + b x_2 + c x_3}{a + b + c}, \frac{a y_1 + b y_2 + c y_3}{a + b + c} \right)
\]
Substituting the values:
- \(a = 25\), \(b = 39\), \(c = 56\)
- \(x_1 = -36\), \(y_1 = 7\)
- \(x_2 = 20\), \(y_2 = 7\)
- \(x_3 = 0\), \(y_3 = -8\)
Calculating the x-coordinate of the incenter:
\[
I_x = \frac{25(-36) + 39(20) + 56(0)}{25 + 39 + 56} = \frac{-900 + 780 + 0}{120} = \frac{-120}{120} = -1
\]
Calculating the y-coordinate of the incenter:
\[
I_y = \frac{25(7) + 39(7) + 56(-8)}{25 + 39 + 56} = \frac{175 + 273 - 448}{120} = \frac{0}{120} = 0
\]
Thus, the coordinates of the incenter are:
\[
I(-1, 0)
\]
### Final Answer
The coordinates of the centroid and incenter of triangle ABC are:
- Centroid: \(\left( -\frac{16}{3}, 2 \right)\)
- Incenter: \((-1, 0)\)
