If the point `3x+4y-24=0` intersects the `X`-axis at the point `A` and the `Y`-axis at the point `B`, then the incentre of the triangle `OAB`, where `O` is the origin, is

Options

A. A. (4, 4)
B. B. (3, 4)
C. C. (2, 2)
D. D. (4, 3)

Explanation

To find the incenter of the triangle OAB formed by the origin O and the points A and B where the line \(3x + 4y - 24 = 0\) intersects the x-axis and y-axis, we will follow these steps:

### Step 1: Find the points of intersection A and B
1. **Find point A (intersection with x-axis)**:
- Set \(y = 0\) in the equation \(3x + 4y - 24 = 0\):
\[ 3x + 4(0) - 24 = 0 \implies 3x = 24 \implies x = 8 \]
- Thus, point A is \((8, 0)\).

2. **Find point B (intersection with y-axis)**:
- Set \(x = 0\) in the equation \(3x + 4y - 24 = 0\):
\[ 3(0) + 4y - 24 = 0 \implies 4y = 24 \implies y = 6 \]
- Thus, point B is \((0, 6)\).

### Step 2: Identify the vertices of triangle OAB
- The vertices of triangle OAB are:
- \(O(0, 0)\)
- \(A(8, 0)\)
- \(B(0, 6)\)

### Step 3: Calculate the lengths of the sides of triangle OAB
1. **Length OA**:
\[ OA = \sqrt{(8 - 0)^2 + (0 - 0)^2} = \sqrt{8^2} = 8 \]

2. **Length OB**:
\[ OB = \sqrt{(0 - 0)^2 + (6 - 0)^2} = \sqrt{6^2} = 6 \]

3. **Length AB**:
\[ AB = \sqrt{(8 - 0)^2 + (0 - 6)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]

### Step 4: Use the incenter formula
The coordinates of the incenter \(I\) of triangle \(OAB\) can be calculated using the formula:
\[ I_x = \frac{aA_x + bB_x + cO_x}{a + b + c} \]
\[ I_y = \frac{aA_y + bB_y + cO_y}{a + b + c} \]
Where:
- \(a = OB = 6\)
- \(b = OA = 8\)
- \(c = AB = 10\)
- \(A(8, 0)\), \(B(0, 6)\), \(O(0, 0)\)

Substituting the values:
\[ I_x = \frac{6 \cdot 8 + 8 \cdot 0 + 10 \cdot 0}{6 + 8 + 10} = \frac{48}{24} = 2 \]
\[ I_y = \frac{6 \cdot 0 + 8 \cdot 6 + 10 \cdot 0}{6 + 8 + 10} = \frac{48}{24} = 2 \]

### Final Result
Thus, the coordinates of the incenter \(I\) are \((2, 2)\).