A straight line through the point `(2,2)` intersects the lines `sqrt(3)x+y=0` and `sqrt(3)x-y=0` at the point `A` and `B ,` respectively. Then find the equation of the line `A B` so that triangle `O A B` is equilateral.

Explanation

To find the equation of the line \( AB \) such that triangle \( OAB \) is equilateral, we will follow these steps:

### Step 1: Identify the lines and their slopes
The given lines are:
1. \( \sqrt{3}x + y = 0 \) (Line 1)
2. \( \sqrt{3}x - y = 0 \) (Line 2)

From these equations, we can rewrite them in slope-intercept form (y = mx + b):
- For Line 1:
\[ y = -\sqrt{3}x \]
The slope \( m_1 = -\sqrt{3} \).

- For Line 2:
\[ y = \sqrt{3}x \]
The slope \( m_2 = \sqrt{3} \).

### Step 2: Find the intersection points A and B
To find points \( A \) and \( B \), we need to determine where a line through the point \( (2, 2) \) intersects these two lines.

Let the equation of the line through \( (2, 2) \) be:
\[ y - 2 = m(x - 2) \]
where \( m \) is the slope of the line.

**Finding point A (intersection with Line 1):**
Substituting \( y = -\sqrt{3}x \) into the line equation:
\[ -\sqrt{3}x - 2 = m(x - 2) \]
This gives:
\[ -\sqrt{3}x - 2 = mx - 2m \]
Rearranging:
\[ (-\sqrt{3} - m)x = -2 + 2m \]
Thus:
\[ x = \frac{-2 + 2m}{-\sqrt{3} - m} \]
Now substitute \( x \) back into \( y = -\sqrt{3}x \) to find \( y_A \).

**Finding point B (intersection with Line 2):**
Substituting \( y = \sqrt{3}x \) into the line equation:
\[ \sqrt{3}x - 2 = m(x - 2) \]
This gives:
\[ \sqrt{3}x - 2 = mx - 2m \]
Rearranging:
\[ (\sqrt{3} - m)x = -2 + 2m \]
Thus:
\[ x = \frac{-2 + 2m}{\sqrt{3} - m} \]
Now substitute \( x \) back into \( y = \sqrt{3}x \) to find \( y_B \).

### Step 3: Ensure triangle OAB is equilateral
For triangle \( OAB \) to be equilateral, the distances \( OA \), \( OB \), and \( AB \) must be equal.

1. Calculate \( OA \) using the distance formula:
\[ OA = \sqrt{(x_A - 0)^2 + (y_A - 0)^2} \]

2. Calculate \( OB \):
\[ OB = \sqrt{(x_B - 0)^2 + (y_B - 0)^2} \]

3. Calculate \( AB \):
\[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \]

Set \( OA = OB = AB \) and solve for \( m \).

### Step 4: Find the equation of line AB
Once we have the coordinates of points \( A \) and \( B \), we can find the slope of line \( AB \):
\[ m_{AB} = \frac{y_B - y_A}{x_B - x_A} \]
Using the point-slope form of the line equation:
\[ y - y_A = m_{AB}(x - x_A) \]
This gives us the equation of line \( AB \).