Two sides of a rhombus OABC ( lying entirely in first quadrant or fourth quadrant) of area equal to 2 sq. units, are `y =x/sqrt(3), y=sqrt(3)x` Then possible coordinates of B is / are ('O' being the origin)

Options

A. `(1+sqrt(3), 1+sqrt(3))`
B. `(-1-sqrt(3), -1-sqrt(3))`
C. `(3+sqrt(3), 3+sqrt(3))`
D. `(sqrt(3)-1, sqrt(3)-1)`

Explanation

To solve the problem, we need to find the possible coordinates of point B in the rhombus OABC, given the area and the equations of two sides of the rhombus.

### Step-by-step Solution:

1. **Identify the Given Information:**
- The area of the rhombus OABC is 2 square units.
- The equations of the sides are \( y = \frac{x}{\sqrt{3}} \) and \( y = \sqrt{3}x \).
- The origin O is at (0,0).

2. **Determine the Angles:**
- The line \( y = \frac{x}{\sqrt{3}} \) has a slope of \( \frac{1}{\sqrt{3}} \), which corresponds to an angle of \( 30^\circ \) with the x-axis.
- The line \( y = \sqrt{3}x \) has a slope of \( \sqrt{3} \), which corresponds to an angle of \( 60^\circ \) with the x-axis.

3. **Calculate the Lengths of the Sides:**
- The area of a rhombus can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times OA \times OC \times \sin(\theta) \]
- Here, \( \theta \) is the angle between OA and OC. Since the angles are \( 30^\circ \) and \( 60^\circ \), the angle between OA and OC is \( 30^\circ \).
- Thus, the area can be expressed as:
\[ 2 = \frac{1}{2} \times OA \times OC \times \sin(30^\circ) \]
- Since \( \sin(30^\circ) = \frac{1}{2} \), we have:
\[ 2 = \frac{1}{2} \times OA \times OC \times \frac{1}{2} \]
- This simplifies to:
\[ OA \times OC = 8 \]

4. **Assuming OA = OC:**
- Let \( OA = OC = x \). Then:
\[ x^2 = 8 \implies x = 2\sqrt{2} \]

5. **Finding Coordinates of Points A and C:**
- The coordinates of point A can be determined from the line \( y = \frac{x}{\sqrt{3}} \):
\[ A = \left(2\sqrt{2} \cos(30^\circ), 2\sqrt{2} \sin(30^\circ)\right) = \left(2\sqrt{2} \cdot \frac{\sqrt{3}}{2}, 2\sqrt{2} \cdot \frac{1}{2}\right) = \left(\sqrt{6}, \sqrt{2}\right) \]
- The coordinates of point C can be determined from the line \( y = \sqrt{3}x \):
\[ C = \left(2\sqrt{2} \cos(60^\circ), 2\sqrt{2} \sin(60^\circ)\right) = \left(2\sqrt{2} \cdot \frac{1}{2}, 2\sqrt{2} \cdot \frac{\sqrt{3}}{2}\right) = \left(\sqrt{2}, \sqrt{6}\right) \]

6. **Calculating the Coordinates of Point B:**
- The angle \( OAB \) is \( 150^\circ \). Using the cosine rule:
\[ OB^2 = OA^2 + AB^2 - 2 \cdot OA \cdot AB \cdot \cos(150^\circ) \]
- Since \( OA = AB = 2\sqrt{2} \):
\[ OB^2 = (2\sqrt{2})^2 + (2\sqrt{2})^2 - 2 \cdot (2\sqrt{2}) \cdot (2\sqrt{2}) \cdot (-\frac{\sqrt{3}}{2}) \]
- Simplifying gives:
\[ OB^2 = 8 + 8 + 8\sqrt{3} = 16 + 8\sqrt{3} \]
- Thus, \( OB = \sqrt{16 + 8\sqrt{3}} \).

7. **Finding Coordinates of B:**
- The coordinates of B can be expressed as:
\[ B = \left(OB \cos(45^\circ), OB \sin(45^\circ)\right) = \left(\frac{\sqrt{16 + 8\sqrt{3}}}{\sqrt{2}}, \frac{\sqrt{16 + 8\sqrt{3}}}{\sqrt{2}}\right) \]
- This simplifies to:
\[ B = \left(\sqrt{8 + 4\sqrt{3}}, \sqrt{8 + 4\sqrt{3}}\right) \]

### Possible Coordinates of B:
The possible coordinates of B are:
- \( \left(1 + \sqrt{3}, 1 + \sqrt{3}\right) \)
- \( \left(-1 - \sqrt{3}, -1 - \sqrt{3}\right) \)