The incentre of the triangle formed by axes and the line `x/a+y/b=1` is

Options

A. `(a/2,b/2)`
B. `(a/3,b/3)`
C. `[(ab)/(a+b+sqrt(a^(2)+b^(2))),(ab)/(a+b+sqrt(a^(2)+b^(2)))]`
D. `[(ab)/(a+b+sqrt(ab)),(ab)/(a+b+sqrt(ab))]`

Answer

To find the incenter of the triangle formed by the axes and the line \( \frac{x}{a} + \frac{y}{b} = 1 \), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The line intersects the x-axis and y-axis at points:
- When \( y = 0 \): \( \frac{x}{a} + 0 = 1 \) → \( x = a \) → Point A is \( (a, 0) \)
- When \( x = 0 \): \( 0 + \frac{y}{b} = 1 \) → \( y = b \) → Point B is \( (0, b) \)
- The third vertex is the origin \( O(0, 0) \).

Thus, the vertices of the triangle are:
- \( O(0, 0) \)
- \( A(a, 0) \)
- \( B(0, b) \)

### Step 2: Calculate the lengths of the sides of the triangle
The lengths of the sides opposite to each vertex are:
- Length \( a \) (opposite to vertex O) = \( b \) (vertical side)
- Length \( b \) (opposite to vertex A) = \( a \) (horizontal side)
- Length \( c \) (opposite to vertex B) = \( \sqrt{a^2 + b^2} \) (hypotenuse)

### Step 3: Use the formula for the incenter
The coordinates of the incenter \( I(x, y) \) can be calculated using the formula:
\[ x = \frac{a \cdot x_1 + b \cdot x_2 + c \cdot x_3}{a + b + c} \]
\[ y = \frac{a \cdot y_1 + b \cdot y_2 + c \cdot y_3}{a + b + c} \]

Where:
- \( (x_1, y_1) = (0, 0) \) (vertex O)
- \( (x_2, y_2) = (a, 0) \) (vertex A)
- \( (x_3, y_3) = (0, b) \) (vertex B)

### Step 4: Substitute the values into the formula
Substituting the coordinates and the lengths:
- For \( x \):
\[ x = \frac{b \cdot 0 + a \cdot a + \sqrt{a^2 + b^2} \cdot 0}{b + a + \sqrt{a^2 + b^2}} = \frac{a^2}{a + b + \sqrt{a^2 + b^2}} \]

- For \( y \):
\[ y = \frac{b \cdot 0 + a \cdot 0 + \sqrt{a^2 + b^2} \cdot b}{b + a + \sqrt{a^2 + b^2}} = \frac{b \sqrt{a^2 + b^2}}{a + b + \sqrt{a^2 + b^2}} \]

### Final Result
Thus, the incenter \( I \) of the triangle formed by the axes and the line \( \frac{x}{a} + \frac{y}{b} = 1 \) is:
\[ I\left(\frac{a^2}{a + b + \sqrt{a^2 + b^2}}, \frac{b \sqrt{a^2 + b^2}}{a + b + \sqrt{a^2 + b^2}}\right) \]