The vertices of a triangle are (0, 0), (1,0) and (0,1). Then excentre opposite to (0, 0) is

Options

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

Answer

To find the excentre opposite to the vertex (0, 0) of the triangle with vertices (0, 0), (1, 0), and (0, 1), we can follow these steps:

### Step 1: Identify the vertices
Let the vertices of the triangle be:
- A(0, 0)
- B(1, 0)
- C(0, 1)

### Step 2: Calculate the lengths of the sides
We need to find the lengths of the sides opposite to each vertex:
- Length of side BC (opposite to A):
\[ BC = \sqrt{(1 - 0)^2 + (0 - 1)^2} = \sqrt{1 + 1} = \sqrt{2} \]
- Length of side AC (opposite to B):
\[ AC = \sqrt{(0 - 0)^2 + (1 - 0)^2} = \sqrt{0 + 1} = 1 \]
- Length of side AB (opposite to C):
\[ AB = \sqrt{(1 - 0)^2 + (0 - 0)^2} = \sqrt{1 + 0} = 1 \]

### Step 3: Assign values to A, B, and C
Let:
- \( A = BC = \sqrt{2} \)
- \( B = AC = 1 \)
- \( C = AB = 1 \)

### Step 4: Use the excentre formula
The coordinates of the excentre opposite to vertex A (0, 0) can be calculated using the formula:
\[ \text{Excentre} = \left( \frac{-A x_1 + B x_2 + C x_3}{-A + B + C}, \frac{-A y_1 + B y_2 + C y_3}{-A + B + C} \right) \]
Substituting the values:
- \( x_1 = 0, y_1 = 0 \) (for A)
- \( x_2 = 1, y_2 = 0 \) (for B)
- \( x_3 = 0, y_3 = 1 \) (for C)

### Step 5: Substitute into the formula
\[ \text{Excentre} = \left( \frac{-\sqrt{2} \cdot 0 + 1 \cdot 1 + 1 \cdot 0}{-\sqrt{2} + 1 + 1}, \frac{-\sqrt{2} \cdot 0 + 1 \cdot 0 + 1 \cdot 1}{-\sqrt{2} + 1 + 1} \right) \]
This simplifies to:
\[ \text{Excentre} = \left( \frac{1}{2 - \sqrt{2}}, \frac{1}{2 - \sqrt{2}} \right) \]

### Step 6: Rationalize the coordinates
To rationalize the coordinates:
\[ \frac{1}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{2 + \sqrt{2}}{(2 - \sqrt{2})(2 + \sqrt{2})} = \frac{2 + \sqrt{2}}{4 - 2} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{\sqrt{2}}{2} \]
Thus, the excentre coordinates are:
\[ \left( 1 + \frac{\sqrt{2}}{2}, 1 + \frac{\sqrt{2}}{2} \right) \]

### Final Answer
The excentre opposite to (0, 0) is:
\[ \left( 1 + \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}} \right) \]