To find the value of \( \lambda \) for which the equation
\[
12x^2 - 10xy + 2y^2 + 11x - 5y + \lambda = 0
\]
represents a pair of straight lines, we will use the condition for a quadratic equation in two variables to represent a pair of straight lines. The condition is given by:
\[
ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0
\]
where \( A \), \( B \), \( C \), \( F \), \( G \), and \( H \) are the coefficients of the equation \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \).
### Step 1: Identify coefficients
From the given equation, we can identify:
- \( A = 12 \)
- \( B = -10 \)
- \( C = 2 \)
- \( D = 11 \)
- \( E = -5 \)
- \( F = \lambda \)
### Step 2: Substitute into the condition
Now we substitute these values into the condition:
\[
12 \cdot 2 \cdot \lambda + 2 \cdot 11 \cdot (-5) \cdot 0 - 12 \cdot 11^2 - (-10)^2 - 2 \cdot (-5)^2 = 0
\]
### Step 3: Calculate each term
Calculating each term:
1. \( ABC = 12 \cdot 2 \cdot \lambda = 24\lambda \)
2. \( 2FGH = 0 \) (since \( H = 0 \))
3. \( -AF^2 = -12 \cdot 11^2 = -12 \cdot 121 = -1452 \)
4. \( -BG^2 = -(-10)^2 = -100 \)
5. \( -CH^2 = -2 \cdot (-5)^2 = -2 \cdot 25 = -50 \)
### Step 4: Combine the terms
Now we combine these terms:
\[
24\lambda - 1452 - 100 - 50 = 0
\]
This simplifies to:
\[
24\lambda - 1602 = 0
\]
### Step 5: Solve for \( \lambda \)
Now, we solve for \( \lambda \):
\[
24\lambda = 1602
\]
\[
\lambda = \frac{1602}{24} = 66.75
\]
### Step 6: Find the equations of the lines
Now, we need to find the equations of the lines. We will rewrite the equation with \( \lambda = 66.75 \):
\[
12x^2 - 10xy + 2y^2 + 11x - 5y + 66.75 = 0
\]
Dividing the entire equation by 2:
\[
6x^2 - 5xy + y^2 + \frac{11}{2}x - \frac{5}{2}y + 33.375 = 0
\]
### Step 7: Use the general form of pair of lines
The general form of the pair of straight lines is:
\[
(y - m_1x - c_1)(y - m_2x - c_2) = 0
\]
From the coefficients, we can derive:
1. \( m_1m_2 = 6 \)
2. \( -m_1 - m_2 = -5 \) (which gives \( m_1 + m_2 = 5 \))
### Step 8: Solve for \( m_1 \) and \( m_2 \)
Let \( m_1 \) and \( m_2 \) be the roots of the quadratic equation \( t^2 - 5t + 6 = 0 \).
Using the quadratic formula:
\[
t = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}
\]
Thus, \( m_1 = 3 \) and \( m_2 = 2 \).
### Step 9: Find \( c_1 \) and \( c_2 \)
Using \( c_1m_2 + c_1m_2 = \frac{11}{2} \):
1. \( 3c_1 + 2c_2 = \frac{11}{2} \)
2. \( c_1 + c_2 = \frac{5}{2} \)
Solving these two equations will yield \( c_1 \) and \( c_2 \).
### Step 10: Find the angle between the lines
The angle \( \theta \) between the two lines is given by:
\[
\tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b}
\]
Substituting \( a = 12 \), \( b = 2 \), and \( h = -10 \):
\[
\tan \theta = \frac{2\sqrt{(-10)^2 - (12)(2)}}{12 + 2} = \frac{2\sqrt{100 - 24}}{14} = \frac{2\sqrt{76}}{14} = \frac{\sqrt{76}}{7}
\]
### Final Result
Thus, the value of \( \lambda \) is \( 66.75 \), the equations of the lines can be derived from the values of \( m_1, m_2, c_1, c_2 \), and the angle between them is given by \( \tan^{-1}\left(\frac{\sqrt{76}}{7}\right) \).
