To find the incentre 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 line \( \frac{x}{a} + \frac{y}{b} = 1 \) intersects the x-axis and y-axis at the points:
- For the x-axis (where \( y = 0 \)):
\[
\frac{x}{a} + \frac{0}{b} = 1 \implies x = a \implies (a, 0)
\]
- For the y-axis (where \( x = 0 \)):
\[
\frac{0}{a} + \frac{y}{b} = 1 \implies y = b \implies (0, b)
\]
- The origin, which is the intersection of the axes, is \( (0, 0) \).
Thus, the vertices of the triangle are \( (0, 0) \), \( (a, 0) \), and \( (0, b) \).
### Step 2: Calculate the lengths of the sides of the triangle
The lengths of the sides opposite to the vertices are:
- Length opposite to vertex \( (0, 0) \) (side \( AB \)):
\[
AB = \sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2}
\]
- Length opposite to vertex \( (a, 0) \) (side \( AC \)):
\[
AC = b
\]
- Length opposite to vertex \( (0, b) \) (side \( BC \)):
\[
BC = a
\]
### Step 3: Use the formula for the coordinates of the incentre
The coordinates of the incentre \( (I_x, I_y) \) can be calculated using the formula:
\[
I_x = \frac{a_1 x_1 + a_2 x_2 + a_3 x_3}{a_1 + a_2 + a_3}
\]
\[
I_y = \frac{a_1 y_1 + a_2 y_2 + a_3 y_3}{a_1 + a_2 + a_3}
\]
where \( a_1, a_2, a_3 \) are the lengths of the sides opposite to the vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \).
Substituting the values:
- \( (x_1, y_1) = (0, 0) \)
- \( (x_2, y_2) = (a, 0) \)
- \( (x_3, y_3) = (0, b) \)
- \( a_1 = \sqrt{a^2 + b^2} \), \( a_2 = b \), \( a_3 = a \)
Calculating \( I_x \):
\[
I_x = \frac{\sqrt{a^2 + b^2} \cdot 0 + b \cdot a + a \cdot 0}{\sqrt{a^2 + b^2} + b + a} = \frac{ab}{\sqrt{a^2 + b^2} + a + b}
\]
Calculating \( I_y \):
\[
I_y = \frac{\sqrt{a^2 + b^2} \cdot 0 + b \cdot 0 + a \cdot b}{\sqrt{a^2 + b^2} + b + a} = \frac{ab}{\sqrt{a^2 + b^2} + a + b}
\]
### Final Result
Thus, the coordinates of the incentre are:
\[
\left( \frac{ab}{\sqrt{a^2 + b^2} + a + b}, \frac{ab}{\sqrt{a^2 + b^2} + a + b} \right)
\]
