Find the incentre of the triangle formed by the points A(7,9), B(3,-7) and C(-3,3).

Answer

To find the incenter of the triangle formed by the points A(7,9), B(3,-7), and C(-3,3), we will follow these steps:

### Step 1: Calculate the lengths of the sides of the triangle

1. **Calculate side \( a \) (BC):**
\[ a = \sqrt{(3 - (-3))^2 + (-7 - 3)^2} = \sqrt{(3 + 3)^2 + (-7 - 3)^2} = \sqrt{6^2 + (-10)^2} = \sqrt{36 + 100} = \sqrt{136} = 2\sqrt{34} \]

2. **Calculate side \( b \) (AC):**
\[ b = \sqrt{(7 - (-3))^2 + (9 - 3)^2} = \sqrt{(7 + 3)^2 + (9 - 3)^2} = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} = 2\sqrt{34} \]

3. **Calculate side \( c \) (AB):**
\[ c = \sqrt{(7 - 3)^2 + (9 - (-7))^2} = \sqrt{(4)^2 + (9 + 7)^2} = \sqrt{16 + 256} = \sqrt{272} = 4\sqrt{17} \]

### Step 2: Use the incenter formula

The coordinates of the incenter \( I(x, y) \) 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 \( I_x \):**
\[ I_x = \frac{(2\sqrt{34}) \cdot 7 + (2\sqrt{34}) \cdot 3 + (4\sqrt{17}) \cdot (-3)}{2\sqrt{34} + 2\sqrt{34} + 4\sqrt{17}} \]
\[ = \frac{14\sqrt{34} + 6\sqrt{34} - 12\sqrt{17}}{4\sqrt{34} + 4\sqrt{17}} = \frac{20\sqrt{34} - 12\sqrt{17}}{4(\sqrt{34} + \sqrt{17})} \]
\[ = \frac{5\sqrt{34} - 3\sqrt{17}}{\sqrt{34} + \sqrt{17}} \]

2. **Calculate \( I_y \):**
\[ I_y = \frac{(2\sqrt{34}) \cdot 9 + (2\sqrt{34}) \cdot (-7) + (4\sqrt{17}) \cdot 3}{2\sqrt{34} + 2\sqrt{34} + 4\sqrt{17}} \]
\[ = \frac{18\sqrt{34} - 14\sqrt{34} + 12\sqrt{17}}{4\sqrt{34} + 4\sqrt{17}} = \frac{4\sqrt{34} + 12\sqrt{17}}{4(\sqrt{34} + \sqrt{17})} \]
\[ = \frac{\sqrt{34} + 3\sqrt{17}}{\sqrt{34} + \sqrt{17}} \]

### Final Incenter Coordinates

Thus, the incenter \( I \) of the triangle is:
\[ I\left(\frac{5\sqrt{34} - 3\sqrt{17}}{\sqrt{34} + \sqrt{17}}, \frac{\sqrt{34} + 3\sqrt{17}}{\sqrt{34} + \sqrt{17}}\right) \]