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. `(sqrt(3)-1)x+(sqrt(3)+1)y=8sqrt(2)`
B. `x+sqrt(3)y=8`
C. `sqrt(3)x+y=8`
D. None of these

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 has a perpendicular from the origin that makes an angle of \( 60^\circ \) with the line \( x + y = 0 \), we can follow these steps:

### Step 1: Understand the Geometry
The line \( L \) makes positive intercepts on the axes, meaning it can be expressed in the intercept form:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
where \( a \) and \( b \) are the x-intercept and y-intercept, respectively.

### Step 2: Determine the Distance from the Origin
The distance \( d \) from the origin to the line can be calculated using the formula:
\[ d = \frac{|c|}{\sqrt{a^2 + b^2}} \]
For our line in intercept form, \( c = -1 \), so the distance becomes:
\[ d = \frac{1}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2}} = 4 \]
This implies:
\[ \sqrt{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{4} \]
Squaring both sides gives:
\[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{16} \]

### Step 3: Relate the Angles
The line \( x + y = 0 \) has a slope of \( -1 \). The angle \( \theta \) that this line makes with the positive x-axis is \( 135^\circ \) (since \( \tan(135^\circ) = -1 \)). The angle \( \alpha \) that the perpendicular from the origin to line \( L \) makes with the x-axis is \( 60^\circ \). Thus:
\[ \theta + \alpha = 135^\circ \]
This implies:
\[ \alpha = 135^\circ - 60^\circ = 75^\circ \]

### Step 4: Write the Equation of Line \( L \)
The equation of the line in terms of the angle \( \alpha \) is given by:
\[ x \cos(75^\circ) + y \sin(75^\circ) = 4 \]

### Step 5: Calculate \( \cos(75^\circ) \) and \( \sin(75^\circ) \)
Using the angle addition formulas:
\[ \cos(75^\circ) = \cos(30^\circ + 45^\circ) = \cos(30^\circ)\cos(45^\circ) - \sin(30^\circ)\sin(45^\circ) \]
Substituting values:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \cos(45^\circ) = \frac{1}{\sqrt{2}}, \quad \sin(30^\circ) = \frac{1}{2}, \quad \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
Calculating:
\[ \cos(75^\circ) = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} - \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

For \( \sin(75^\circ) \):
\[ \sin(75^\circ) = \sin(30^\circ + 45^\circ) = \sin(30^\circ)\cos(45^\circ) + \cos(30^\circ)\sin(45^\circ) \]
Calculating:
\[ \sin(75^\circ) = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1 + \sqrt{3}}{2\sqrt{2}} \]

### Step 6: Substitute into the Line Equation
Substituting back into the line equation:
\[ x \frac{\sqrt{3} - 1}{2\sqrt{2}} + y \frac{1 + \sqrt{3}}{2\sqrt{2}} = 4 \]
Multiplying through by \( 2\sqrt{2} \):
\[ (\sqrt{3} - 1)x + (1 + \sqrt{3})y = 8\sqrt{2} \]

### Final Equation
The equation of line \( L \) is:
\[ (\sqrt{3} - 1)x + (1 + \sqrt{3})y = 8\sqrt{2} \]