Consider a triangle ABC with at `(0, -3), (-2sqrt(3), 3)` and `(2sqrt(3), 3)` respectively. The incentre of the triangle with vertices at the mid points of the sides of triangle ABC is

Options

A. (0, 0)
B. (0, 1)
C. `(-sqrt(3), -sqrt(3))`
D. `(sqrt(3),sqrt(3))`

Answer

To find the incenter of the triangle formed by the midpoints of the sides of triangle ABC with vertices A(0, -3), B(-2√3, 3), and C(2√3, 3), we will follow these steps:

### Step 1: Find the midpoints of the sides of triangle ABC

1. **Midpoint D (between A and B)**:
\[ D = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) = \left( \frac{0 + (-2\sqrt{3})}{2}, \frac{-3 + 3}{2} \right) = \left( -\sqrt{3}, 0 \right) \]

2. **Midpoint E (between B and C)**:
\[ E = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) = \left( \frac{-2\sqrt{3} + 2\sqrt{3}}{2}, \frac{3 + 3}{2} \right) = \left( 0, 3 \right) \]

3. **Midpoint F (between A and C)**:
\[ F = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{0 + 2\sqrt{3}}{2}, \frac{-3 + 3}{2} \right) = \left( \sqrt{3}, 0 \right) \]

### Step 2: Calculate the lengths of the sides of triangle DEF

1. **Length DE**:
\[ DE = \sqrt{(0 - (-\sqrt{3}))^2 + (3 - 0)^2} = \sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3} \]

2. **Length EF**:
\[ EF = \sqrt{(\sqrt{3} - 0)^2 + (0 - 3)^2} = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3} \]

3. **Length FD**:
\[ FD = \sqrt{(-\sqrt{3} - \sqrt{3})^2 + (0 - 0)^2} = \sqrt{(-2\sqrt{3})^2} = 2\sqrt{3} \]

### Step 3: Calculate the incenter coordinates

The incenter (I) of triangle DEF can be calculated using the formula:
\[ I_x = \frac{a \cdot x_A + b \cdot x_B + c \cdot x_C}{a + b + c} \]
\[ I_y = \frac{a \cdot y_A + b \cdot y_B + c \cdot y_C}{a + b + c} \]
Where \(a\), \(b\), and \(c\) are the lengths of sides opposite to vertices A, B, and C respectively.

Substituting the values:
- \(a = EF = 2\sqrt{3}\)
- \(b = FD = 2\sqrt{3}\)
- \(c = DE = 2\sqrt{3}\)

1. **Calculate \(I_x\)**:
\[ I_x = \frac{(2\sqrt{3})(-\sqrt{3}) + (2\sqrt{3})(0) + (2\sqrt{3})(\sqrt{3})}{2\sqrt{3} + 2\sqrt{3} + 2\sqrt{3}} = \frac{-6 + 6}{6\sqrt{3}} = \frac{0}{6\sqrt{3}} = 0 \]

2. **Calculate \(I_y\)**:
\[ I_y = \frac{(2\sqrt{3})(0) + (2\sqrt{3})(3) + (2\sqrt{3})(0)}{2\sqrt{3} + 2\sqrt{3} + 2\sqrt{3}} = \frac{0 + 6\sqrt{3}}{6\sqrt{3}} = 1 \]

### Final Result
The coordinates of the incenter of triangle DEF are:
\[ \text{Incenter} = (0, 1) \]