If (0, 1), (1, 1) and (1, 0) are the mid points of the sides of a triangle, the coordinates of its incentre are

Options

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

Answer

To find the coordinates of the incentre of a triangle given the midpoints of its sides, we can follow these steps:

### Step 1: Identify the midpoints
The midpoints of the triangle's sides are given as:
- Midpoint M1 = (0, 1)
- Midpoint M2 = (1, 1)
- Midpoint M3 = (1, 0)

### Step 2: Set up the equations for the vertices
Let the vertices of the triangle be A(x1, y1), B(x2, y2), and C(x3, y3). The midpoints can be expressed in terms of the vertices:
- M1 = ((x1 + x2)/2, (y1 + y2)/2) = (0, 1)
- M2 = ((x2 + x3)/2, (y2 + y3)/2) = (1, 1)
- M3 = ((x3 + x1)/2, (y3 + y1)/2) = (1, 0)

### Step 3: Solve for the vertices
From the first midpoint equation:
1. \( \frac{x1 + x2}{2} = 0 \) → \( x1 + x2 = 0 \) → \( x2 = -x1 \)
2. \( \frac{y1 + y2}{2} = 1 \) → \( y1 + y2 = 2 \) → \( y2 = 2 - y1 \)

From the second midpoint equation:
3. \( \frac{x2 + x3}{2} = 1 \) → \( x2 + x3 = 2 \) → \( x3 = 2 - x2 \)
4. \( \frac{y2 + y3}{2} = 1 \) → \( y2 + y3 = 2 \) → \( y3 = 2 - y2 \)

From the third midpoint equation:
5. \( \frac{x3 + x1}{2} = 1 \) → \( x3 + x1 = 2 \) → \( x1 = 2 - x3 \)
6. \( \frac{y3 + y1}{2} = 0 \) → \( y3 + y1 = 0 \) → \( y1 = -y3 \)

### Step 4: Substitute and solve
Substituting \( x2 = -x1 \) into equation (3):
- \( x3 = 2 - (-x1) = 2 + x1 \)

Now substituting \( y2 = 2 - y1 \) into equation (4):
- \( y3 = 2 - (2 - y1) = y1 \)

Now substituting \( y3 = y1 \) into equation (6):
- \( y1 = -y1 \) → \( 2y1 = 0 \) → \( y1 = 0 \)

Substituting \( y1 = 0 \) into \( y2 = 2 - y1 \):
- \( y2 = 2 - 0 = 2 \)

Now substituting \( y1 = 0 \) into \( y3 = y1 \):
- \( y3 = 0 \)

Now substituting \( y1 = 0 \) into \( x2 = -x1 \):
- \( x1 + (-x1) = 0 \) → \( x1 = 0 \)

Now substituting \( x1 = 0 \) into \( x3 = 2 + x1 \):
- \( x3 = 2 + 0 = 2 \)

So we have:
- A(0, 0)
- B(0, 2)
- C(2, 0)

### Step 5: Calculate side lengths
Now we calculate the lengths of the sides:
- \( a = BC = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \)
- \( b = AC = \sqrt{(0 - 0)^2 + (0 - 2)^2} = \sqrt{4} = 2 \)
- \( c = AB = \sqrt{(0 - 2)^2 + (0 - 0)^2} = \sqrt{4} = 2 \)

### Step 6: Use the incenter formula
The coordinates of the incentre (I) are given by:
\[ I_x = \frac{a \cdot x1 + b \cdot x2 + c \cdot x3}{a + b + c} \]
\[ I_y = \frac{a \cdot y1 + b \cdot y2 + c \cdot y3}{a + b + c} \]

Substituting the values:
\[ I_x = \frac{(2\sqrt{2} \cdot 0) + (2 \cdot 0) + (2 \cdot 2)}{2\sqrt{2} + 2 + 2} = \frac{0 + 0 + 4}{2\sqrt{2} + 4} \]
\[ I_y = \frac{(2\sqrt{2} \cdot 0) + (2 \cdot 2) + (2 \cdot 0)}{2\sqrt{2} + 2 + 2} = \frac{0 + 4 + 0}{2\sqrt{2} + 4} \]

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

### Step 7: Simplify the coordinates
We can factor out 2 from the numerator and denominator:
\[ I = \left( \frac{2}{\sqrt{2} + 2}, \frac{2}{\sqrt{2} + 2} \right) \]

### Final Answer
The coordinates of the incentre are:
\[ I = \left( 2 - \sqrt{2}, 2 - \sqrt{2} \right) \]