A straight line of length 9 units slides with ends A, B always on x and y axes respectiv Locus of centroid of AOAB is

Options

A. `x^(2)+y^(2)=3`
B. `x^(2)+y^(2)=9`
C. `x^(2)+y^(2)=1`
D. `x^(2)+y^(2)=81`

Answer

To find the locus of the centroid of triangle OAB, where O is the origin (0,0), A is on the x-axis, and B is on the y-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Define Points A and B**:
Let point A be at (a, 0) on the x-axis and point B be at (0, b) on the y-axis.

2. **Centroid Formula**:
The centroid (G) of triangle OAB is given by the formula:
\[ G\left(h, k\right) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
For triangle OAB, the coordinates of points are:
- O(0, 0)
- A(a, 0)
- B(0, b)

Thus, the coordinates of the centroid G are:
\[ h = \frac{0 + a + 0}{3} = \frac{a}{3} \]
\[ k = \frac{0 + 0 + b}{3} = \frac{b}{3} \]

3. **Express a and b in terms of h and k**:
From the equations for h and k, we can express a and b as:
\[ a = 3h \]
\[ b = 3k \]

4. **Length of the line segment AB**:
The length of line segment AB is given as 9 units. The distance formula between points A and B is:
\[ \text{Distance} = \sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} \]
Setting this equal to 9 gives us:
\[ \sqrt{a^2 + b^2} = 9 \]

5. **Square both sides**:
Squaring both sides, we have:
\[ a^2 + b^2 = 81 \]

6. **Substitute a and b**:
Substitute \(a = 3h\) and \(b = 3k\) into the equation:
\[ (3h)^2 + (3k)^2 = 81 \]
This simplifies to:
\[ 9h^2 + 9k^2 = 81 \]

7. **Divide by 9**:
Dividing the entire equation by 9 gives:
\[ h^2 + k^2 = 9 \]

8. **Locus of the Centroid**:
The equation \(h^2 + k^2 = 9\) represents a circle with radius 3 centered at the origin. Therefore, the locus of the centroid G(h, k) is:
\[ x^2 + y^2 = 9 \]

### Conclusion:
The locus of the centroid of triangle OAB is given by the equation:
\[ x^2 + y^2 = 9 \]