If O(0,0), A(4,0) and B(0,3) are the vertice of a triangle OAB, then the coordinates of the excentre oppsite to the vertex O(0,0)

Answer

To find the coordinates of the excentre opposite to the vertex O(0,0) of triangle OAB with vertices O(0,0), A(4,0), and B(0,3), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the vertices of the triangle**:
- O(0,0)
- A(4,0)
- B(0,3)

2. **Calculate the lengths of the sides of the triangle**:
- Let \( a \) be the length opposite to vertex O (side AB).
- Let \( b \) be the length opposite to vertex A (side OB).
- Let \( c \) be the length opposite to vertex B (side OA).

Using the distance formula:
- \( a = \text{distance between A and B} = \sqrt{(4-0)^2 + (0-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)
- \( b = \text{distance between O and B} = \sqrt{(0-0)^2 + (0-3)^2} = \sqrt{9} = 3 \)
- \( c = \text{distance between O and A} = \sqrt{(0-4)^2 + (0-0)^2} = \sqrt{16} = 4 \)

3. **Use the excentre formula**:
The coordinates of the excentre opposite to vertex O can be calculated using the formula:
\[ I_O = \left( \frac{aX_A + bX_B - cX_O}{a+b+c}, \frac{aY_A + bY_B - cY_O}{a+b+c} \right) \]
where:
- \( (X_O, Y_O) \) are the coordinates of O,
- \( (X_A, Y_A) \) are the coordinates of A,
- \( (X_B, Y_B) \) are the coordinates of B.

Substituting the values:
- \( X_A = 4, Y_A = 0 \)
- \( X_B = 0, Y_B = 3 \)
- \( X_O = 0, Y_O = 0 \)

Plugging in the values:
\[ I_O = \left( \frac{5 \cdot 4 + 3 \cdot 0 - 4 \cdot 0}{5 + 3 + 4}, \frac{5 \cdot 0 + 3 \cdot 3 - 4 \cdot 0}{5 + 3 + 4} \right) \]

4. **Calculate the coordinates**:
- For the x-coordinate:
\[ I_O^x = \frac{20 + 0 - 0}{12} = \frac{20}{12} = \frac{5}{3} \]
- For the y-coordinate:
\[ I_O^y = \frac{0 + 9 - 0}{12} = \frac{9}{12} = \frac{3}{4} \]

5. **Final coordinates of the excentre**:
Thus, the coordinates of the excentre opposite to vertex O(0,0) are:
\[ I_O = \left( \frac{5}{3}, \frac{3}{4} \right) \]