The equations of lines passing through the point (1,0) and at distance of `(sqrt(3))/(2)` units from the origin are (i) `sqrt(3)x+y-sqrt(3)=0,sqrt(3)x-ysqrt(3)=0` (ii) `sqrt(3)x+y+sqrt(3)=0,sqrt(3)x-y+sqrt(3)=0` (iii) `x+sqrt(3)y-sqrt(3)=0,x-sqrt(3)y-sqrt(3)=0` (iv) none of these

Options

A. `sqrt(3)x+y-sqrt(3)=0,sqrt(3)x-ysqrt(3)=0`
B. `sqrt(3)x+y+sqrt(3)=0,sqrt(3)x-y+sqrt(3)=0`
C. `x+sqrt(3)y-sqrt(3)=0,x-sqrt(3)y-sqrt(3)=0`
D. none of these

Answer

To solve the problem, we need to find the equations of lines that pass through the point (1, 0) and are at a distance of \(\frac{\sqrt{3}}{2}\) units from the origin.

### Step-by-Step Solution:

1. **Identify the Point and Distance**:
- The point through which the lines pass is \( (1, 0) \).
- The distance from the origin (0, 0) to the line is given as \( \frac{\sqrt{3}}{2} \).

2. **Equation of the Line**:
- Let the slope of the line be \( m \). The equation of the line in point-slope form is:
\[ y - 0 = m(x - 1) \implies y = mx - m \]
- Rearranging gives:
\[ mx - y - m = 0 \]

3. **Distance from the Origin to the Line**:
- The formula for the distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
- Here, \( A = m \), \( B = -1 \), and \( C = -m \). The distance from the origin (0, 0) to the line is:
\[ \frac{|m(0) - 1(0) - m|}{\sqrt{m^2 + (-1)^2}} = \frac{| - m |}{\sqrt{m^2 + 1}} = \frac{m}{\sqrt{m^2 + 1}} \]
- Setting this equal to \( \frac{\sqrt{3}}{2} \):
\[ \frac{m}{\sqrt{m^2 + 1}} = \frac{\sqrt{3}}{2} \]

4. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 2m = \sqrt{3} \sqrt{m^2 + 1} \]
- Squaring both sides results in:
\[ 4m^2 = 3(m^2 + 1) \implies 4m^2 = 3m^2 + 3 \]
- Rearranging gives:
\[ m^2 = 3 \implies m = \pm \sqrt{3} \]

5. **Finding the Equations of the Lines**:
- For \( m = \sqrt{3} \):
\[ y = \sqrt{3}(x - 1) \implies y = \sqrt{3}x - \sqrt{3} \implies \sqrt{3}x - y - \sqrt{3} = 0 \]
- For \( m = -\sqrt{3} \):
\[ y = -\sqrt{3}(x - 1) \implies y = -\sqrt{3}x + \sqrt{3} \implies \sqrt{3}x + y - \sqrt{3} = 0 \]

6. **Final Equations**:
- The equations of the lines are:
\[ \sqrt{3}x - y - \sqrt{3} = 0 \quad \text{and} \quad \sqrt{3}x + y - \sqrt{3} = 0 \]

### Conclusion:
The correct option is (ii) \( \sqrt{3}x + y + \sqrt{3} = 0, \sqrt{3}x - y + \sqrt{3} = 0 \).