The equation of the line whose perpendicular distance from the origin is 3 units and the angle which the normal makes with with the positive direction of x -axis is `30^(@)` is

Options

A. `x+sqrt(3)y=3`
B. `sqrt(3)x+y=6`
C. `sqrt(3)x+y=1`
D. `x+sqrt(3)y=6`

Answer

To find the equation of the line whose perpendicular distance from the origin is 3 units and the angle which the normal makes with the positive direction of the x-axis is 30 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between the normal and the line:**
The normal to the line makes an angle of 30 degrees with the positive x-axis. The slope of the normal can be calculated using the tangent of the angle:
\[ m_1 = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

2. **Find the slope of the line:**
The slope of the line (denoted as \(m\)) is the negative reciprocal of the slope of the normal:
\[ m = -\frac{1}{m_1} = -\sqrt{3} \]

3. **Write the equation of the line in slope-intercept form:**
The equation of the line can be expressed as:
\[ y = mx + c \]
Substituting the value of \(m\):
\[ y = -\sqrt{3}x + c \]

4. **Use the distance formula to find \(c\):**
The perpendicular distance \(d\) from the origin (0, 0) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line, rearranging \(y = -\sqrt{3}x + c\) gives:
\[ \sqrt{3}x + y - c = 0 \]
Here, \(A = \sqrt{3}\), \(B = 1\), and \(C = -c\).

Setting \(d = 3\) (the given distance from the origin):
\[ 3 = \frac{|\sqrt{3}(0) + 1(0) - c|}{\sqrt{(\sqrt{3})^2 + (1)^2}} = \frac{| -c |}{\sqrt{3 + 1}} = \frac{| -c |}{2} \]
This simplifies to:
\[ 3 = \frac{|c|}{2} \]
Thus, we find:
\[ |c| = 6 \]
Therefore, \(c\) can be either \(6\) or \(-6\).

5. **Write the equations of the lines:**
Using \(c = 6\):
\[ y = -\sqrt{3}x + 6 \]
Rearranging gives:
\[ \sqrt{3}x + y - 6 = 0 \]
Using \(c = -6\):
\[ y = -\sqrt{3}x - 6 \]
Rearranging gives:
\[ \sqrt{3}x + y + 6 = 0 \]

6. **Final answer:**
The equation of the line can be either:
\[ \sqrt{3}x + y - 6 = 0 \quad \text{or} \quad \sqrt{3}x + y + 6 = 0 \]