Find the coordinates of incentre of the triangle whose are (4, -2), (-2, 4) and (5, 5).

Answer

To find the coordinates of the incenter of the triangle with vertices at A(4, -2), B(-2, 4), and C(5, 5), we will follow these steps:

### Step 1: Calculate the lengths of the sides of the triangle
We will denote the lengths of the sides opposite to vertices A, B, and C as a, b, and c respectively.

1. **Calculate side a (BC)**:
\[ a = \sqrt{(x_B - x_C)^2 + (y_B - y_C)^2} = \sqrt{((-2) - 5)^2 + (4 - 5)^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \]

2. **Calculate side b (AC)**:
\[ b = \sqrt{(x_A - x_C)^2 + (y_A - y_C)^2} = \sqrt{(4 - 5)^2 + ((-2) - 5)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \]

3. **Calculate side c (AB)**:
\[ c = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2} = \sqrt{(4 - (-2))^2 + ((-2) - 4)^2} = \sqrt{(6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]

### Step 2: Use the incenter formula
The coordinates of the incenter (I) can be calculated using the formula:
\[ I_x = \frac{a \cdot x_A + b \cdot x_B + c \cdot x_C}{a + b + c} \]
\[ I_y = \frac{a \cdot y_A + b \cdot y_B + c \cdot y_C}{a + b + c} \]

### Step 3: Substitute the values into the formula
1. **Calculate \(I_x\)**:
\[ I_x = \frac{(5\sqrt{2}) \cdot 4 + (5\sqrt{2}) \cdot (-2) + (6\sqrt{2}) \cdot 5}{5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} \]
\[ = \frac{20\sqrt{2} - 10\sqrt{2} + 30\sqrt{2}}{16\sqrt{2}} = \frac{40\sqrt{2}}{16\sqrt{2}} = \frac{40}{16} = \frac{5}{2} \]

2. **Calculate \(I_y\)**:
\[ I_y = \frac{(5\sqrt{2}) \cdot (-2) + (5\sqrt{2}) \cdot 4 + (6\sqrt{2}) \cdot 5}{5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} \]
\[ = \frac{-10\sqrt{2} + 20\sqrt{2} + 30\sqrt{2}}{16\sqrt{2}} = \frac{40\sqrt{2}}{16\sqrt{2}} = \frac{40}{16} = \frac{5}{2} \]

### Step 4: Conclusion
The coordinates of the incenter of the triangle are:
\[ \left(\frac{5}{2}, \frac{5}{2}\right) \]