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

Options

A. `(a/b,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 given by the equation \( \frac{x}{a} + \frac{y}{b} = 1 \), we can follow these steps:

### Step 1: Identify the vertices of the triangle
The triangle is formed by:
- The x-axis (where \( y = 0 \))
- The y-axis (where \( x = 0 \))
- The line \( \frac{x}{a} + \frac{y}{b} = 1 \)

To find the vertices:
1. **Intersection with the x-axis**: Set \( y = 0 \) in the line equation:
\[ \frac{x}{a} + 0 = 1 \implies x = a \implies (a, 0) \]

2. **Intersection with the y-axis**: Set \( x = 0 \) in the line equation:
\[ 0 + \frac{y}{b} = 1 \implies y = b \implies (0, b) \]

3. **Origin**: The third vertex is the origin, which is \( (0, 0) \).

Thus, the vertices of the triangle are \( (a, 0) \), \( (0, b) \), and \( (0, 0) \).

### Step 2: Calculate the lengths of the sides of the triangle
Let:
- \( A = (0, b) \)
- \( B = (a, 0) \)
- \( C = (0, 0) \)

The lengths of the sides opposite to these vertices are:
- Side opposite to \( A \) (length \( BC \)): \( \sqrt{a^2 + b^2} \)
- Side opposite to \( B \) (length \( AC \)): \( b \)
- Side opposite to \( C \) (length \( AB \)): \( a \)

### Step 3: Use the incenter formula
The incenter \( I \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \) and corresponding opposite side lengths \( a \), \( b \), and \( c \) is given by:
\[ I_x = \frac{a x_1 + b x_2 + c x_3}{a + b + c}, \quad I_y = \frac{a y_1 + b y_2 + c y_3}{a + b + c} \]

### Step 4: Substitute the values
Using the vertices:
- \( A(0, b) \)
- \( B(a, 0) \)
- \( C(0, 0) \)

And the side lengths:
- \( a = \sqrt{a^2 + b^2} \)
- \( b = b \)
- \( c = a \)

Substituting into the incenter formula:
1. For \( I_x \):
\[ I_x = \frac{\sqrt{a^2 + b^2} \cdot 0 + b \cdot a + a \cdot 0}{\sqrt{a^2 + b^2} + b + a} = \frac{ba}{\sqrt{a^2 + b^2} + b + a} \]

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

### Step 5: Final coordinates of the incenter
Thus, the coordinates of the incenter \( I \) are:
\[ I = \left( \frac{ab}{\sqrt{a^2 + b^2} + b + a}, \frac{b\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2} + b + a} \right) \]