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

Options

A. `2+sqrt2`
B. `2-sqrt2`
C. `1+sqrt2`
D. `1-sqrt2`

Answer

To find the x-coordinate of the incenter of the triangle with midpoints of its sides given as (0, 1), (1, 1), and (1, 0), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The midpoints of the sides of the triangle are given. We can denote the midpoints as:
- D (0, 1)
- E (1, 1)
- F (1, 0)

To find the vertices A, B, and C of the triangle, we can use the midpoint formula. The midpoint M of a line segment connecting points (x1, y1) and (x2, y2) is given by:
\[ M = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2} \right) \]

### Step 2: Calculate the vertices
Using the midpoints, we can find the vertices of the triangle:
- Let A be the vertex opposite to D, B opposite to E, and C opposite to F.

1. For midpoint D (0, 1):
\[ A = (x1, y1) \quad \text{and} \quad C = (0, 0) \quad \text{(assuming C is at the origin)} \]
Thus, using the midpoint formula:
\[ D = \left( \frac{x1 + 0}{2}, \frac{y1 + 0}{2} \right) \Rightarrow (0, 1) = \left( \frac{x1}{2}, \frac{y1}{2} \right) \]
From this, we get:
\[ x1 = 0 \quad \text{and} \quad y1 = 2 \quad \Rightarrow A = (0, 2) \]

2. For midpoint E (1, 1):
\[ B = (x2, y2) \quad \text{and} \quad C = (0, 0) \]
Thus:
\[ E = \left( \frac{x2 + 0}{2}, \frac{y2 + 0}{2} \right) \Rightarrow (1, 1) = \left( \frac{x2}{2}, \frac{y2}{2} \right) \]
From this, we get:
\[ x2 = 2 \quad \text{and} \quad y2 = 2 \quad \Rightarrow B = (2, 2) \]

3. For midpoint F (1, 0):
\[ C = (x3, y3) \quad \text{and} \quad A = (0, 2) \]
Thus:
\[ F = \left( \frac{x3 + 0}{2}, \frac{y3 + 2}{2} \right) \Rightarrow (1, 0) = \left( \frac{x3}{2}, \frac{y3 + 2}{2} \right) \]
From this, we get:
\[ x3 = 2 \quad \text{and} \quad y3 = -2 \quad \Rightarrow C = (2, -2) \]

### Step 3: Calculate the lengths of the sides
Now we have the vertices:
- A (0, 2)
- B (2, 2)
- C (2, 0)

Next, we calculate the lengths of the sides:
- AB:
\[ AB = \sqrt{(2 - 0)^2 + (2 - 2)^2} = \sqrt{4} = 2 \]
- BC:
\[ BC = \sqrt{(2 - 2)^2 + (2 - 0)^2} = \sqrt{4} = 2 \]
- CA:
\[ CA = \sqrt{(2 - 0)^2 + (0 - 2)^2} = \sqrt{8} = 2\sqrt{2} \]

### Step 4: Calculate the coordinates of the incenter
The x-coordinate of the incenter (I) can be calculated using the formula:
\[ x_I = \frac{a \cdot x_A + b \cdot x_B + c \cdot x_C}{a + b + c} \]
where:
- a = length of BC = 2
- b = length of CA = 2√2
- c = length of AB = 2

Substituting the values:
\[ x_I = \frac{2 \cdot 0 + 2\sqrt{2} \cdot 2 + 2 \cdot 2}{2 + 2\sqrt{2} + 2} \]
\[ = \frac{0 + 4\sqrt{2} + 4}{4 + 2\sqrt{2}} \]
\[ = \frac{4 + 4\sqrt{2}}{4 + 2\sqrt{2}} \]

### Step 5: Simplify the expression
To simplify:
\[ = \frac{4(1 + \sqrt{2})}{2(2 + \sqrt{2})} = \frac{2(1 + \sqrt{2})}{2 + \sqrt{2}} \]

### Final Answer
The x-coordinate of the incenter is:
\[ x_I = 2(1 + \sqrt{2})/(2 + \sqrt{2}) \]