If `DeltaOAB` is an equilateral triangle (O is the origin and A is a point on the x - axis) , then centroid of the traignle will be
(a point P (x,y) is said to be rational if both x and y are rational )

Options

A. always rational
B. rotional if B is rational
C. rational if A is rational
D. never rational

Answer

To solve the problem, we need to find the coordinates of the centroid of the equilateral triangle \( \Delta OAB \) where \( O \) is the origin, \( A \) is a point on the x-axis, and \( B \) is another point determined by the properties of the equilateral triangle.

### Step-by-Step Solution:

1. **Define Points**:
- Let \( O \) be the origin, so \( O(0, 0) \).
- Let \( A \) be a point on the x-axis, say \( A(a, 0) \) where \( a \) is a positive real number.
- Since \( \Delta OAB \) is an equilateral triangle, we need to find the coordinates of point \( B \).

2. **Coordinates of Point B**:
- The distance \( OA \) is \( a \).
- The angle \( AOB \) is \( 60^\circ \). We can find the coordinates of \( B \) using trigonometry:
- The coordinates of \( B \) can be expressed as:
\[ B\left(a \cos 60^\circ, a \sin 60^\circ\right) = \left(a \cdot \frac{1}{2}, a \cdot \frac{\sqrt{3}}{2}\right) = \left(\frac{a}{2}, \frac{a\sqrt{3}}{2}\right) \]

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

- Plugging in the coordinates:
\[ G\left(\frac{0 + a + \frac{a}{2}}{3}, \frac{0 + 0 + \frac{a\sqrt{3}}{2}}{3}\right) \]
\[ G\left(\frac{a + \frac{a}{2}}{3}, \frac{\frac{a\sqrt{3}}{2}}{3}\right) = \left(\frac{\frac{3a}{2}}{3}, \frac{a\sqrt{3}}{6}\right) = \left(\frac{a}{2}, \frac{a\sqrt{3}}{6}\right) \]

4. **Rationality of the Centroid**:
- The x-coordinate of the centroid is \( \frac{a}{2} \), which is rational if \( a \) is rational.
- The y-coordinate of the centroid is \( \frac{a\sqrt{3}}{6} \). Since \( \sqrt{3} \) is irrational, \( \frac{a\sqrt{3}}{6} \) will be irrational unless \( a = 0 \).

5. **Conclusion**:
- Therefore, the centroid \( G \) will not be a rational point unless \( a = 0 \) (which is not a valid case since \( A \) must be a point on the x-axis).
- Hence, the centroid will **never be rational**.

### Final Answer:
The centroid of triangle \( \Delta OAB \) will **never be rational**.