Statement 1: If in a triangle, orthocentre, circumcentre and centroid are rational points, then its vertices must also be rational points.
Statement : 2 If the vertices of a triangle are rational points, then the centroid, circumcentre and orthocentre are also rational points.

Options

A. Statement 1 is true, Statement 2 is true and Statement 2 is correct explanation for Statement 1.
B. Statement 1 is true , Statement 2 is true and Statement 2 is not the correct exlpanation for Statement 1.
C. Statement 1 is true, Statement 2 is false.
D. Statement 1 is false, Statement 2 is true.

Explanation

To solve the given problem, we need to analyze both statements regarding the properties of triangles in the coordinate system.

### Step-by-Step Solution:

1. **Understanding Statement 1**:
- Statement 1 claims that if the orthocenter, circumcenter, and centroid of a triangle are rational points, then the vertices of the triangle must also be rational points.
- We need to determine if this statement is always true or if there are counterexamples.

2. **Understanding Statement 2**:
- Statement 2 states that if the vertices of a triangle are rational points, then the centroid, circumcenter, and orthocenter are also rational points.
- This statement is generally true. We can prove this by using the formulas for the centroid, circumcenter, and orthocenter.

3. **Proving Statement 2**:
- Let the vertices of the triangle be \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \), where \( x_1, y_1, x_2, y_2, x_3, y_3 \) are all rational numbers.
- The **centroid** \( G \) is given by:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Since the sum and division of rational numbers yield rational numbers, \( G \) is rational.
- The **circumcenter** and **orthocenter** can also be shown to be rational using the properties of the triangle and the coordinates of its vertices. Thus, Statement 2 is true.

4. **Finding a Counterexample for Statement 1**:
- To disprove Statement 1, we need to find a case where the orthocenter, circumcenter, and centroid are rational, but the vertices are not.
- Consider the vertices:
- \( A(1, 0) \)
- \( B\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)
- \( C\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \)
- The centroid \( G \) can be calculated as:
\[ G = \left( \frac{1 - \frac{1}{2} - \frac{1}{2}}{3}, \frac{0 + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}}{3} \right) = (0, 0) \]
- The circumcenter and orthocenter can also be shown to be rational through calculations.
- However, the vertices \( B \) and \( C \) contain \( \sqrt{3} \), which is irrational.

5. **Conclusion**:
- Statement 1 is false, as we found a counterexample.
- Statement 2 is true, as we proved it holds for rational vertices.

### Final Answer:
- Statement 1 is false.
- Statement 2 is true.