If the equation `12x^(2)-10xy+2y^(2)+11x-5y+lamda=0` represents a pair of straight lines then `lamda=`

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To find the value of \(\lambda\) such that the equation \(12x^2 - 10xy + 2y^2 + 11x - 5y + \lambda = 0\) represents a pair of straight lines, we can follow these steps:

### Step 1: Identify coefficients
We can rewrite the given equation in the standard form of a conic section:
\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]
From the given equation, we identify:
- \(a = 12\)
- \(h = -5\) (since \(2h = -10\))
- \(b = 2\)
- \(g = \frac{11}{2}\) (since \(2g = 11\))
- \(f = -\frac{5}{2}\) (since \(2f = -5\))
- \(c = \lambda\)

### Step 2: Use the condition for a pair of straight lines
For the equation to represent a pair of straight lines, the determinant must be zero:
\[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]
Substituting the values we identified:
\[ \begin{vmatrix} 12 & -5 & \frac{11}{2} \\ -5 & 2 & -\frac{5}{2} \\ \frac{11}{2} & -\frac{5}{2} & \lambda \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant:
\[ = 12 \left(2\lambda - \left(-\frac{5}{2}\right)\left(-\frac{5}{2}\right)\right) - (-5)\left(-5\lambda + \frac{11}{2}\left(-\frac{5}{2}\right)\right) + \frac{11}{2}\left(-5\lambda + 2\left(-\frac{5}{2}\right)\right) \]

Calculating each term:
1. First term:
\[ 12 \left(2\lambda - \frac{25}{4}\right) = 24\lambda - 75 \]
2. Second term:
\[ -5 \left(-5\lambda - \frac{55}{4}\right) = 25\lambda + \frac{275}{4} \]
3. Third term:
\[ \frac{11}{2} \left(-5\lambda - 5\right) = -\frac{55}{2}\lambda - \frac{55}{2} \]

Combining these:
\[ 24\lambda - 75 + 25\lambda + \frac{275}{4} - \frac{55}{2}\lambda - \frac{55}{2} = 0 \]

### Step 4: Simplify the equation
Combine like terms:
\[ (24 + 25 - \frac{55}{2})\lambda - 75 + \frac{275}{4} - \frac{110}{4} = 0 \]
Calculating the coefficients:
\[ 49\lambda - 75 + \frac{165}{4} = 0 \]
Convert \(75\) to a fraction:
\[ 49\lambda - \frac{300}{4} + \frac{165}{4} = 0 \]
Combine:
\[ 49\lambda - \frac{135}{4} = 0 \]
Thus:
\[ 49\lambda = \frac{135}{4} \]
\[ \lambda = \frac{135}{4 \times 49} = \frac{135}{196} \]

### Final Answer
\[ \lambda = \frac{135}{196} \]