The line `3x+4y-24=0` cuts the x -axis at A and y-axis at B. Then the increntre of a triangle OAB, where O is the origin is

Options

A. (1,2)
B. (2,2)
C. (12,2)
D. (2,12)

Answer

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

### Step 1: Find the intercepts A and B
1. **Find the x-intercept (A)**:
- Set \(y = 0\) in the equation \(3x + 4y - 24 = 0\).
- This gives us \(3x - 24 = 0\) or \(3x = 24\).
- Therefore, \(x = 8\).
- So, the coordinates of point A are \(A(8, 0)\).

2. **Find the y-intercept (B)**:
- Set \(x = 0\) in the equation \(3x + 4y - 24 = 0\).
- This gives us \(4y - 24 = 0\) or \(4y = 24\).
- Therefore, \(y = 6\).
- So, the coordinates of point B are \(B(0, 6)\).

### Step 2: Identify the coordinates of points O, A, and B
- The coordinates of the points are:
- \(O(0, 0)\)
- \(A(8, 0)\)
- \(B(0, 6)\)

### Step 3: Calculate the lengths of the sides of triangle OAB
1. **Length of OA**:
- The distance from O to A is \(OA = 8\) units.

2. **Length of OB**:
- The distance from O to B is \(OB = 6\) units.

3. **Length of AB**:
- Use the distance formula to find \(AB\):
\[ AB = \sqrt{(8 - 0)^2 + (0 - 6)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ units}. \]

### Step 4: Use the incenter formula
The coordinates of the incenter \(I\) of triangle OAB can be calculated using the formula:
\[ I = \left( \frac{aX_A + bX_B + cX_O}{a + b + c}, \frac{aY_A + bY_B + cY_O}{a + b + c} \right) \]
where:
- \(a = AB = 10\)
- \(b = OA = 8\)
- \(c = OB = 6\)
- \(X_A = 8, Y_A = 0\)
- \(X_B = 0, Y_B = 6\)
- \(X_O = 0, Y_O = 0\)

Substituting the values:
\[ I_x = \frac{10 \cdot 8 + 8 \cdot 0 + 6 \cdot 0}{10 + 8 + 6} = \frac{80}{24} = \frac{10}{3} \]
\[ I_y = \frac{10 \cdot 0 + 8 \cdot 6 + 6 \cdot 0}{10 + 8 + 6} = \frac{48}{24} = 2 \]

### Final Result
Thus, the coordinates of the incenter \(I\) are:
\[ I\left(\frac{10}{3}, 2\right) \]