Find the in-centre of the right angled isosceles triangle having one vertex at the origin and having the other two vertices at (6, 0) and (0, 6).

Options

A. `3(2+sqrt2),3(2+sqrt(2))`
B. `3(2-sqrt(2)),3(2-sqrt(2))`
C. `2+sqrt(2),2+sqrt(2)`
D. `2-sqrt(2),2-sqrt(2)`

Answer

To find the in-center of the right-angled isosceles triangle with vertices at (0, 0), (6, 0), and (0, 6), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Vertices**:
The vertices of the triangle are:
- A (0, 0)
- B (6, 0)
- C (0, 6)

2. **Calculate the Lengths of the Sides**:
- Length of AB (base) = 6 units (from (0, 0) to (6, 0))
- Length of AC (height) = 6 units (from (0, 0) to (0, 6))
- Length of BC (hypotenuse) = √[(6-0)² + (0-6)²] = √[36 + 36] = √72 = 6√2 units

3. **Identify the Sides**:
- Perpendicular side (p) = 6 (AC)
- Base (b) = 6 (AB)
- Hypotenuse (h) = 6√2 (BC)

4. **Calculate the Inradius (r)**:
The formula for the inradius \( r \) of a right triangle is given by:
\[ r = \frac{p + b - h}{2} \]
Substituting the values:
\[ r = \frac{6 + 6 - 6\sqrt{2}}{2} \]
Simplifying:
\[ r = \frac{12 - 6\sqrt{2}}{2} = 6 - 3\sqrt{2} \]

5. **Determine the Coordinates of the In-center**:
For a right-angled triangle, the in-center coordinates can be expressed as \( (r, r) \):
\[ \text{In-center} = (6 - 3\sqrt{2}, 6 - 3\sqrt{2}) \]

### Final Answer:
The in-center of the triangle is at the coordinates:
\[ (6 - 3\sqrt{2}, 6 - 3\sqrt{2}) \]