The incentre of a triangle with vertices `(7, 1),(-1, 5)` and `(3+2sqrt(3),3+4sqrt(3))` is

Options

A. `(3+(2)/(sqrt(3)),3+(4)/(sqrt(3)))`
B. `(1+(2)/(3sqrt(3)),1+(4)/(3sqrt(3)))`
C. `(7,1)`
D. None of these

Explanation

To find the incenter of the triangle with vertices \( A(7, 1) \), \( B(-1, 5) \), and \( C(3 + 2\sqrt{3}, 3 + 4\sqrt{3}) \), we will follow these steps:

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

1. **Calculate side \( a \)** (opposite vertex \( A \)):
\[ a = BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} \]
Substituting the coordinates:
\[ a = \sqrt{(3 + 2\sqrt{3} - (-1))^2 + (3 + 4\sqrt{3} - 5)^2} \]
Simplifying:
\[ a = \sqrt{(4 + 2\sqrt{3})^2 + (4\sqrt{3} - 2)^2} \]
Expanding the squares:
\[ a = \sqrt{(16 + 16\sqrt{3} + 12) + (48 - 16\sqrt{3} + 4)} \]
\[ a = \sqrt{80} = 4\sqrt{5} \]

2. **Calculate side \( b \)** (opposite vertex \( B \)):
\[ b = AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} \]
Substituting the coordinates:
\[ b = \sqrt{(3 + 2\sqrt{3} - 7)^2 + (3 + 4\sqrt{3} - 1)^2} \]
Simplifying:
\[ b = \sqrt{(2\sqrt{3} - 4)^2 + (4\sqrt{3} + 2)^2} \]
Expanding the squares:
\[ b = \sqrt{(12 - 16\sqrt{3} + 16) + (48 + 16\sqrt{3} + 4)} \]
\[ b = \sqrt{80} = 4\sqrt{5} \]

3. **Calculate side \( c \)** (opposite vertex \( C \)):
\[ c = AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \]
Substituting the coordinates:
\[ c = \sqrt{(-1 - 7)^2 + (5 - 1)^2} \]
Simplifying:
\[ c = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \]

### Step 2: Use the incenter formula

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

Substituting the values we found:
- \( a = 4\sqrt{5} \)
- \( b = 4\sqrt{5} \)
- \( c = 4\sqrt{5} \)

Calculating \( I_x \):
\[ I_x = \frac{4\sqrt{5} \cdot 7 + 4\sqrt{5} \cdot (-1) + 4\sqrt{5} \cdot (3 + 2\sqrt{3})}{4\sqrt{5} + 4\sqrt{5} + 4\sqrt{5}} \]
\[ = \frac{4\sqrt{5} (7 - 1 + 3 + 2\sqrt{3})}{12\sqrt{5}} = \frac{4\sqrt{5} (9 + 2\sqrt{3})}{12\sqrt{5}} = \frac{9 + 2\sqrt{3}}{3} \]

Calculating \( I_y \):
\[ I_y = \frac{4\sqrt{5} \cdot 1 + 4\sqrt{5} \cdot 5 + 4\sqrt{5} \cdot (3 + 4\sqrt{3})}{4\sqrt{5} + 4\sqrt{5} + 4\sqrt{5}} \]
\[ = \frac{4\sqrt{5} (1 + 5 + 3 + 4\sqrt{3})}{12\sqrt{5}} = \frac{4\sqrt{5} (9 + 4\sqrt{3})}{12\sqrt{5}} = \frac{9 + 4\sqrt{3}}{3} \]

### Final Result

Thus, the coordinates of the incenter \( I \) are:
\[ I\left(\frac{9 + 2\sqrt{3}}{3}, \frac{9 + 4\sqrt{3}}{3}\right) \]