Consider the line passing through `(sqrt 3,1) ` and ` (1,sqrt3) ` Then number of rational points lying on the line is (i) `1` (ii) atleast `2` (iii) `infty` (iv) none of these

Options

A. 1
B. at most 2
C. `oo`
D. none of these

Answer

To determine the number of rational points lying on the line passing through the points \((\sqrt{3}, 1)\) and \((1, \sqrt{3})\), we will follow these steps:

### Step 1: Find the slope of the line
The slope \(m\) of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For our points \((\sqrt{3}, 1)\) and \((1, \sqrt{3})\):
\[ m = \frac{\sqrt{3} - 1}{1 - \sqrt{3}} \]

### Step 2: Simplify the slope
To simplify the slope:
\[ m = \frac{\sqrt{3} - 1}{1 - \sqrt{3}} = \frac{\sqrt{3} - 1}{-(\sqrt{3} - 1)} = -1 \]
Thus, the slope of the line is \(-1\).

### Step 3: Find the equation of the line
Using the point-slope form of the equation of a line \(y - y_1 = m(x - x_1)\), we can use point \((\sqrt{3}, 1)\):
\[ y - 1 = -1(x - \sqrt{3}) \]
This simplifies to:
\[ y = -x + \sqrt{3} + 1 \]
or
\[ y = -x + (\sqrt{3} + 1) \]

### Step 4: Identify rational points on the line
A rational point \((x, y)\) on this line must satisfy the equation with both \(x\) and \(y\) being rational numbers.

### Step 5: Analyze the points
Since \(\sqrt{3}\) is irrational, \(\sqrt{3} + 1\) is also irrational. Therefore, for \(y\) to be rational, \(x\) must also be irrational because \(y = -x + (\sqrt{3} + 1)\) implies that if \(x\) is rational, \(y\) would be irrational.

### Conclusion
Since both points \((\sqrt{3}, 1)\) and \((1, \sqrt{3})\) are irrational, and the line connecting them does not yield any rational points, we conclude that there are no rational points lying on this line.

Thus, the answer is:
**(iv) none of these**.