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

### Step 2: Calculate the lengths of the sides of the triangle
Using the distance formula, we can find the lengths of the sides opposite to each vertex:
- Length of side BC (denoted as a):
\[ a = \sqrt{(1 - 0)^2 + (0 - 1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

- Length of side AC (denoted as b):
\[ b = \sqrt{(0 - 0)^2 + (1 - 0)^2} = \sqrt{0 + 1} = 1 \]

- Length of side AB (denoted as c):
\[ c = \sqrt{(1 - 0)^2 + (0 - 0)^2} = \sqrt{1 + 0} = 1 \]

### Step 3: Use the formula for the excentre coordinates
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:
- \(x_1 = 0\), \(y_1 = 0\)
- \(x_2 = 1\), \(y_2 = 0\)
- \(x_3 = 0\), \(y_3 = 1\)
- \(a = \sqrt{2}\), \(b = 1\), \(c = 1\)

### Step 4: Calculate the x-coordinate of the excentre
\[ x = \frac{-\sqrt{2} \cdot 0 + 1 \cdot 1 + 1 \cdot 0}{-\sqrt{2} + 1 + 1} = \frac{1}{2 - \sqrt{2}} \]

### Step 5: Calculate the y-coordinate of the excentre
\[ y = \frac{-\sqrt{2} \cdot 0 + 1 \cdot 0 + 1 \cdot 1}{-\sqrt{2} + 1 + 1} = \frac{1}{2 - \sqrt{2}} \]

### Step 6: Rationalize the denominators
To rationalize the denominator:
\[ x = \frac{1}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{2 + \sqrt{2}}{4 - 2} = 1 + \frac{1}{\sqrt{2}} \]
Similarly for y:
\[ y = \frac{1}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{2 + \sqrt{2}}{4 - 2} = 1 + \frac{1}{\sqrt{2}} \]

### Final Result
Thus, the coordinates of the excentre opposite to (0, 0) are:
\[ \left( 1 + \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}} \right) \]