If the coordinates of vertices of a triangle is always rational then the triangle cannot be

Options

A. Scalene
B. Isosceles
C. Rightangle
D. Equilateral

Answer

To solve the problem, we need to analyze the area of a triangle whose vertices have rational coordinates.

### Step-by-Step Solution:

1. **Understanding the Coordinates**:
Let the vertices of the triangle be \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \). Given that all coordinates \( x_1, y_1, x_2, y_2, x_3, y_3 \) are rational numbers.

**Hint**: Remember that rational numbers can be expressed as the ratio of two integers.

2. **Area of the Triangle**:
The area \( A \) of the triangle formed by these vertices can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Since \( x_1, y_1, x_2, y_2, x_3, y_3 \) are all rational, the expression inside the absolute value will also be rational.

**Hint**: The area formula involves rational operations (addition, subtraction, multiplication) which will yield a rational result if all inputs are rational.

3. **Rational Area Calculation**:
Thus, the area \( A \) is a rational number because it is half of a rational number, which is still rational.

**Hint**: Dividing a rational number by 2 still results in a rational number.

4. **Considering Special Cases**:
Now, consider the case of an equilateral triangle. The area of an equilateral triangle with side length \( a \) is given by:
\[ A = \frac{\sqrt{3}}{4} a^2 \]
Here, \( \sqrt{3} \) is an irrational number. If \( a \) is rational, then \( a^2 \) is rational, but multiplying by \( \sqrt{3} \) results in an irrational area.

**Hint**: Identify that the presence of \( \sqrt{3} \) in the area formula indicates that the area cannot be rational if the triangle is equilateral.

5. **Conclusion**:
Since we established that the area of a triangle with rational vertices is rational, and the area of an equilateral triangle is irrational when its sides are rational, we conclude that a triangle with all rational vertices cannot be an equilateral triangle.

**Final Answer**: The triangle cannot be an equilateral triangle.