The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1), (1, 1) and (1, 0) is

Options

A. `2+sqrt(2)`
B. `2-sqrt(2)`
C. `1+sqrt(2)`
D. `1-sqrt(2)`

Explanation

To find the x-coordinate of the incenter of the triangle whose midpoints of the sides are given as (0, 1), (1, 1), and (1, 0), we will follow a systematic approach.

### Step 1: Identify the Midpoints
The midpoints of the sides of the triangle are given as:
- \( D(0, 1) \)
- \( E(1, 1) \)
- \( F(1, 0) \)

### Step 2: Find the Vertices of the Triangle
To find the vertices \( A \), \( B \), and \( C \) of the triangle, we will use the property that the coordinates of the vertices can be calculated by:
- \( A = E + F - D \)
- \( B = D + F - E \)
- \( C = D + E - F \)

#### Finding Vertex A:
\[ A = (1, 1) + (1, 0) - (0, 1) = (1 + 1 - 0, 1 + 0 - 1) = (2, 0) \]

#### Finding Vertex B:
\[ B = (0, 1) + (1, 0) - (1, 1) = (0 + 1 - 1, 1 + 0 - 1) = (0, 0) \]

#### Finding Vertex C:
\[ C = (0, 1) + (1, 1) - (1, 0) = (0 + 1 - 1, 1 + 1 - 0) = (0, 2) \]

### Step 3: Calculate the Lengths of the Sides
Next, we need to calculate the lengths of the sides \( AB \), \( BC \), and \( CA \).

#### Length of AB:
\[ AB = \sqrt{(2 - 0)^2 + (0 - 0)^2} = \sqrt{4} = 2 \]

#### Length of BC:
\[ BC = \sqrt{(0 - 0)^2 + (0 - 2)^2} = \sqrt{4} = 2 \]

#### Length of CA:
\[ CA = \sqrt{(2 - 0)^2 + (0 - 2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 4: Calculate the Coordinates of the Incenter
The coordinates of the incenter \( I \) can be calculated using the formula:
\[ I_x = \frac{aA_x + bB_x + cC_x}{a + b + c} \]
\[ I_y = \frac{aA_y + bB_y + cC_y}{a + b + c} \]
where \( a, b, c \) are the lengths of the sides opposite to vertices \( A, B, C \) respectively.

Substituting the values:
- \( a = BC = 2 \)
- \( b = CA = 2\sqrt{2} \)
- \( c = AB = 2 \)

Calculating \( I_x \):
\[ I_x = \frac{2 \cdot 2 + 2\sqrt{2} \cdot 0 + 2 \cdot 0}{2 + 2\sqrt{2} + 2} = \frac{4}{4 + 2\sqrt{2}} \]
Simplifying:
\[ I_x = \frac{4}{2(2 + \sqrt{2})} = \frac{2}{2 + \sqrt{2}} \]
Rationalizing the denominator:
\[ I_x = \frac{2(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{4 - 2\sqrt{2}}{4 - 2} = 2 - \sqrt{2} \]

### Final Answer
The x-coordinate of the incenter of the triangle is:
\[ \boxed{2 - \sqrt{2}} \]