Find the incentre of the triangle, whose vertices are A(3,2) B(7,2) and C(7,5).

Answer

To find the incenter of the triangle with vertices A(3,2), B(7,2), and C(7,5), we will follow these steps:

### Step 1: Identify the vertices
The vertices of the triangle are:
- A(3, 2)
- B(7, 2)
- C(7, 5)

### Step 2: Calculate the lengths of the sides
We will use the distance formula to find the lengths of the sides opposite to each vertex. The distance formula is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

1. **Length of side BC (opposite to A)**:
\[ a = BC = \sqrt{(7 - 7)^2 + (5 - 2)^2} = \sqrt{0 + 9} = 3 \]

2. **Length of side AC (opposite to B)**:
\[ b = AC = \sqrt{(7 - 3)^2 + (5 - 2)^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

3. **Length of side AB (opposite to C)**:
\[ c = AB = \sqrt{(7 - 3)^2 + (2 - 2)^2} = \sqrt{(4)^2 + 0} = \sqrt{16} = 4 \]

### Step 3: Use the incenter formula
The coordinates of the incenter (I) can be calculated using the formula:
\[ I_x = \frac{a x_1 + b x_2 + c x_3}{a + b + c} \]
\[ I_y = \frac{a y_1 + b y_2 + c y_3}{a + b + c} \]

Substituting the values:
- \(x_1 = 3\), \(y_1 = 2\) (for A)
- \(x_2 = 7\), \(y_2 = 2\) (for B)
- \(x_3 = 7\), \(y_3 = 5\) (for C)

### Step 4: Calculate \(I_x\)
\[ I_x = \frac{3 \cdot 3 + 5 \cdot 7 + 4 \cdot 7}{3 + 5 + 4} = \frac{9 + 35 + 28}{12} = \frac{72}{12} = 6 \]

### Step 5: Calculate \(I_y\)
\[ I_y = \frac{3 \cdot 2 + 5 \cdot 2 + 4 \cdot 5}{3 + 5 + 4} = \frac{6 + 10 + 20}{12} = \frac{36}{12} = 3 \]

### Step 6: Conclusion
The incenter of the triangle is:
\[ I(6, 3) \]