If `m_(1)` and `m_(2)` are the roots of an equation `x^(2)+(sqrt3+2)x+(sqrt3-1)=0`, then the area of the triangle formed by the lines `y=m_(1)x,y=m_(2)x,y=c` is

Options

A. `((sqrt33+sqrt11)/(4))c^(2)`
B. `((sqrt32+sqrt11)/(16))c`
C. `((sqrt33+sqrt10)/(4))c^(2)`
D. `((sqrt33+sqrt21)/(4))c^(3)`

Answer

To solve the problem, we need to find the area of the triangle formed by the lines \( y = m_1 x \), \( y = m_2 x \), and \( y = c \), where \( m_1 \) and \( m_2 \) are the roots of the quadratic equation \( x^2 + (\sqrt{3} + 2)x + (\sqrt{3} - 1) = 0 \).

### Step 1: Find the roots \( m_1 \) and \( m_2 \)

The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:
\[ m_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = \sqrt{3} + 2 \), and \( c = \sqrt{3} - 1 \).

Calculating \( b^2 - 4ac \):
\[ b^2 = (\sqrt{3} + 2)^2 = 3 + 4\sqrt{3} + 4 = 7 + 4\sqrt{3} \]
\[ 4ac = 4 \cdot 1 \cdot (\sqrt{3} - 1) = 4\sqrt{3} - 4 \]
Now, substituting these into the discriminant:
\[ b^2 - 4ac = (7 + 4\sqrt{3}) - (4\sqrt{3} - 4) = 7 + 4 = 11 \]

Now, substituting back into the quadratic formula:
\[ m_{1,2} = \frac{-(\sqrt{3} + 2) \pm \sqrt{11}}{2} \]
Thus, we have:
\[ m_1 = \frac{-(\sqrt{3} + 2) - \sqrt{11}}{2}, \quad m_2 = \frac{-(\sqrt{3} + 2) + \sqrt{11}}{2} \]

### Step 2: Calculate the area of the triangle

The area \( A \) of the triangle formed by the lines \( y = m_1 x \), \( y = m_2 x \), and \( y = c \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the height is \( c \) and the base is the distance between the two lines \( y = m_1 x \) and \( y = m_2 x \) at \( y = c \).

The intersection points of the lines with \( y = c \) are:
\[ x_1 = \frac{c}{m_1}, \quad x_2 = \frac{c}{m_2} \]
The base of the triangle is:
\[ \text{base} = |x_1 - x_2| = \left| \frac{c}{m_1} - \frac{c}{m_2} \right| = c \left| \frac{1}{m_1} - \frac{1}{m_2} \right| \]
Using the formula for the difference of reciprocals:
\[ \frac{1}{m_1} - \frac{1}{m_2} = \frac{m_2 - m_1}{m_1 m_2} \]
Now, we need to find \( m_1 m_2 \):
\[ m_1 m_2 = c = \sqrt{3} - 1 \quad \text{(from Vieta's formulas)} \]

Now, we can calculate \( m_2 - m_1 \):
\[ m_2 - m_1 = \sqrt{11} \]

Thus, the base becomes:
\[ \text{base} = c \cdot \frac{\sqrt{11}}{m_1 m_2} = c \cdot \frac{\sqrt{11}}{\sqrt{3} - 1} \]

### Step 3: Putting it all together

Now substituting back into the area formula:
\[ A = \frac{1}{2} \times c \cdot \frac{\sqrt{11}}{\sqrt{3} - 1} \times c = \frac{c^2 \sqrt{11}}{2(\sqrt{3} - 1)} \]

### Final Answer

The area of the triangle formed by the lines is:
\[ A = \frac{c^2 \sqrt{11}}{2(\sqrt{3} - 1)} \]