A straight line `L` at a distance of `4` units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of `60^(@)` with the line `x+y=0`. Then, an equation of the line `L` is

Options

A. `x+sqrt(3)y=8`
B. `(sqrt(3)+1)x+(sqrt(3)-1)y=8sqrt(2)`
C. `sqrt(3)x+y=8`
D. `(sqrt(3)-1)x+(sqrt(3)+1)y=8sqrt(2)`

Explanation

To find the equation of the line \( L \) that is at a distance of 4 units from the origin, makes positive intercepts on the coordinate axes, and forms a 60-degree angle with the perpendicular from the origin to the line \( x + y = 0 \), we can follow these steps:

### Step 1: Determine the slope of the line \( x + y = 0 \)
The line \( x + y = 0 \) can be rewritten in slope-intercept form as \( y = -x \). The slope \( m_1 \) of this line is:
\[ m_1 = -1 \]

### Step 2: Find the angle corresponding to the slope
The angle \( \theta_1 \) that this line makes with the positive x-axis can be calculated using:
\[ \tan(\theta_1) = m_1 = -1 \]
Thus, \( \theta_1 = 135^\circ \) (since the tangent is negative in the second quadrant).

### Step 3: Calculate the angle \( \alpha \) of line \( L \)
Given that the perpendicular from the origin to line \( L \) makes a \( 60^\circ \) angle with the line \( x + y = 0 \):
\[ \theta_1 + 60^\circ + \alpha = 180^\circ \]
Substituting \( \theta_1 = 135^\circ \):
\[ 135^\circ + 60^\circ + \alpha = 180^\circ \]
\[ \alpha = 180^\circ - 195^\circ = -15^\circ \]
Since angles can be expressed positively, we can use \( \alpha = 75^\circ \) (as angles are considered modulo \( 180^\circ \)).

### Step 4: Use the normal form of the line
The normal form of the line can be expressed as:
\[ x \cos(\alpha) + y \sin(\alpha) = p \]
where \( p \) is the distance from the origin (4 units) and \( \alpha = 75^\circ \).

### Step 5: Calculate \( \cos(75^\circ) \) and \( \sin(75^\circ) \)
Using the angle addition formulas:
\[ \cos(75^\circ) = \cos(45^\circ + 30^\circ) = \cos(45^\circ)\cos(30^\circ) - \sin(45^\circ)\sin(30^\circ) \]
\[ = \left(\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\cdot\frac{1}{2}\right) = \frac{\sqrt{6} - \sqrt{2}}{4} \]

\[ \sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ) \]
\[ = \left(\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\cdot\frac{1}{2}\right) = \frac{\sqrt{6} + \sqrt{2}}{4} \]

### Step 6: Substitute values into the normal form
Substituting \( \cos(75^\circ) \), \( \sin(75^\circ) \), and \( p = 4 \):
\[ x \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) + y \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = 4 \]
Multiplying through by 4 to eliminate the fraction:
\[ x(\sqrt{6} - \sqrt{2}) + y(\sqrt{6} + \sqrt{2}) = 16 \]

### Step 7: Rearranging the equation
Rearranging gives the final equation of the line \( L \):
\[ (\sqrt{6} - \sqrt{2})x + (\sqrt{6} + \sqrt{2})y = 16 \]

### Step 8: Finalizing the equation
This can be expressed in the standard form, but the equation we derived is sufficient for the problem.