A straight line through the point (2, 2) intersects the lines `sqrt(3)x+y=0` and `sqrt(3)x-y=0` at the points A and B. The equation of the line AB so that `DeltaOAB` is equilateral, is:

Options

A. `x-2=0`
B. `y-2=0`
C. `x+y-4=0`
D. None of these

Answer

To solve the problem step by step, we need to find the equation of the line that passes through the point (2, 2) and intersects the lines \( \sqrt{3}x + y = 0 \) and \( \sqrt{3}x - y = 0 \) at points A and B such that triangle OAB is equilateral.

### Step 1: Understand the given lines
The lines given are:
1. \( \sqrt{3}x + y = 0 \) (which can be rewritten as \( y = -\sqrt{3}x \))
2. \( \sqrt{3}x - y = 0 \) (which can be rewritten as \( y = \sqrt{3}x \))

These lines intersect the x-axis at angles of \( 60^\circ \) and \( 120^\circ \) respectively.

### Step 2: Determine the slopes of the lines
The slopes of the lines are:
- For \( y = -\sqrt{3}x \), the slope \( m_1 = -\sqrt{3} \).
- For \( y = \sqrt{3}x \), the slope \( m_2 = \sqrt{3} \).

### Step 3: Find the angle between the two lines
The angle between the two lines can be calculated using the formula for the angle between two lines:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the values:
\[ \tan(\theta) = \left| \frac{-\sqrt{3} - \sqrt{3}}{1 + (-\sqrt{3})(\sqrt{3})} \right| = \left| \frac{-2\sqrt{3}}{1 - 3} \right| = \left| \frac{-2\sqrt{3}}{-2} \right| = \sqrt{3} \]
This implies that \( \theta = 60^\circ \).

### Step 4: Find the angle for the equilateral triangle
For triangle OAB to be equilateral, the angles at O, A, and B must all be \( 60^\circ \). The angle between the two lines is \( 60^\circ \), so the angle of the line through point (2, 2) must also be \( 60^\circ \).

### Step 5: Determine the slope of the line through point (2, 2)
The line that makes an angle of \( 60^\circ \) with the positive x-axis has a slope of:
\[ m = \tan(60^\circ) = \sqrt{3} \]

### Step 6: Write the equation of the line through (2, 2)
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (2, 2) \) and \( m = \sqrt{3} \):
\[ y - 2 = \sqrt{3}(x - 2) \]
Expanding this:
\[ y - 2 = \sqrt{3}x - 2\sqrt{3} \]
Rearranging gives:
\[ y = \sqrt{3}x - 2\sqrt{3} + 2 \]

### Step 7: Finalize the equation
The equation of the line can be rewritten in standard form:
\[ y - \sqrt{3}x + (2 - 2\sqrt{3}) = 0 \]
This is the required equation of the line AB.

### Conclusion
The equation of the line AB such that triangle OAB is equilateral is:
\[ y = \sqrt{3}x - 2\sqrt{3} + 2 \]