If `m_(1),m_(2)` be the roots of the equation `x^(2)+(sqrt(3)+2)x+sqrt(3)-1 =0`, then the area of the triangle formed by the lines `y = m_(1)x,y = m_(2)x` and `y = 2` is

Options

A. `sqrt(33)-sqrt(11)` sq. units
B. `sqrt(11) +sqrt(33)` sq. units
C. `2sqrt(33)` sq. units
D. 121 sq. units

Explanation

To find the area of the triangle formed by the lines \(y = m_1 x\), \(y = m_2 x\), and \(y = 2\), we will follow these steps:

### Step 1: Find the roots \(m_1\) and \(m_2\) of the quadratic equation
The given equation is:
\[ x^2 + (\sqrt{3} + 2)x + (\sqrt{3} - 1) = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = \sqrt{3} + 2\), and \(c = \sqrt{3} - 1\):

1. Calculate the discriminant:
\[ D = b^2 - 4ac = (\sqrt{3} + 2)^2 - 4 \cdot 1 \cdot (\sqrt{3} - 1) \]
Expanding this gives:
\[ D = (3 + 4\sqrt{3} + 4) - (4\sqrt{3} - 4) = 7 + 4 \]
Thus, \(D = 11\).

2. Now, substitute back into the quadratic formula:
\[ m_1, m_2 = \frac{-(\sqrt{3} + 2) \pm \sqrt{11}}{2} \]

### Step 2: Determine the coordinates of the triangle vertices
The lines intersect at the following points:

1. **Point A** (intersection of \(y = m_1 x\) and \(y = 2\)):
\[ 2 = m_1 x \Rightarrow x = \frac{2}{m_1} \Rightarrow A\left(\frac{2}{m_1}, 2\right) \]

2. **Point B** (intersection of \(y = m_2 x\) and \(y = 2\)):
\[ 2 = m_2 x \Rightarrow x = \frac{2}{m_2} \Rightarrow B\left(\frac{2}{m_2}, 2\right) \]

3. **Point C** (intersection of \(y = m_1 x\) and \(y = m_2 x\)):
\[ m_1 x = m_2 x \Rightarrow x = 0 \Rightarrow C(0, 0) \]

### Step 3: Calculate the area of the triangle
The area \(A\) of triangle formed by points \(A\), \(B\), and \(C\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ x_1 = \frac{2}{m_1}, y_1 = 2, x_2 = \frac{2}{m_2}, y_2 = 2, x_3 = 0, y_3 = 0 \]
The area becomes:
\[ \text{Area} = \frac{1}{2} \left| \frac{2}{m_1}(2 - 0) + \frac{2}{m_2}(0 - 2) + 0(2 - 2) \right| \]
This simplifies to:
\[ = \frac{1}{2} \left| \frac{4}{m_1} - \frac{4}{m_2} \right| = 2 \left| \frac{1}{m_1} - \frac{1}{m_2} \right| \]

### Step 4: Express \(\left| \frac{1}{m_1} - \frac{1}{m_2} \right|\)
Using the relationship between roots:
\[ \left| \frac{1}{m_1} - \frac{1}{m_2} \right| = \frac{|m_2 - m_1|}{m_1 m_2} \]
From the quadratic roots, we know:
- \(m_1 + m_2 = -(\sqrt{3} + 2)\)
- \(m_1 m_2 = \sqrt{3} - 1\)

### Step 5: Calculate \(|m_2 - m_1|\)
Using the formula:
\[ |m_2 - m_1| = \sqrt{(m_1 + m_2)^2 - 4m_1 m_2} \]
Substituting the values:
\[ = \sqrt{(-\sqrt{3} - 2)^2 - 4(\sqrt{3} - 1)} = \sqrt{(3 + 4\sqrt{3} + 4) - (4\sqrt{3} - 4)} = \sqrt{11} \]

### Step 6: Final area calculation
Substituting back:
\[ \text{Area} = 2 \cdot \frac{\sqrt{11}}{\sqrt{3} - 1} \]
Rationalizing:
\[ = 2 \cdot \frac{\sqrt{11}(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = 2 \cdot \frac{\sqrt{11}(\sqrt{3} + 1)}{2} = \sqrt{11}(\sqrt{3} + 1) \]

### Conclusion
Thus, the area of the triangle is:
\[ \text{Area} = \sqrt{33} + \sqrt{11} \text{ square units.} \]