The value of `lambda` with `|lambda| lt 16` such that `2x^(2) - 10xy +12y^(2) +5x +lambda y - 3 = 0` represents a pair of straight lines is

Options

A. `-10`
B. `-9`
C. 10
D. 9

Explanation

To solve the problem, we need to find the value of \( \lambda \) such that the equation

\[ 2x^2 - 10xy + 12y^2 + 5x + \lambda y - 3 = 0 \]

represents a pair of straight lines. For this to happen, the determinant \( \Delta \) must be equal to zero.

### Step 1: Identify coefficients
We can compare the given equation with the general form of the conic section:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

From the given equation, we can identify the coefficients:
- \( a = 2 \)
- \( h = -5 \)
- \( b = 12 \)
- \( g = \frac{5}{2} \)
- \( f = \frac{\lambda}{2} \)
- \( c = -3 \)

### Step 2: Write the determinant condition
The condition for the equation to represent a pair of straight lines is given by:

\[ \Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]

Substituting the values we identified:

\[ \Delta = \begin{vmatrix} 2 & -5 & \frac{5}{2} \\ -5 & 12 & \frac{\lambda}{2} \\ \frac{5}{2} & \frac{\lambda}{2} & -3 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we use the formula for a 3x3 determinant:

\[ \Delta = a \begin{vmatrix} b & f \\ f & c \end{vmatrix} - h \begin{vmatrix} h & f \\ g & c \end{vmatrix} + g \begin{vmatrix} h & b \\ g & f \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 12 & \frac{\lambda}{2} \\ \frac{\lambda}{2} & -3 \end{vmatrix} = 12(-3) - \left(\frac{\lambda}{2}\right)^2 = -36 - \frac{\lambda^2}{4} \)

2. \( \begin{vmatrix} -5 & \frac{\lambda}{2} \\ \frac{5}{2} & -3 \end{vmatrix} = (-5)(-3) - \left(\frac{\lambda}{2}\right)\left(\frac{5}{2}\right) = 15 - \frac{5\lambda}{4} \)

3. \( \begin{vmatrix} -5 & 12 \\ \frac{5}{2} & \frac{\lambda}{2} \end{vmatrix} = (-5)\left(\frac{\lambda}{2}\right) - (12)\left(\frac{5}{2}\right) = -\frac{5\lambda}{2} - 30 \)

Now substituting back into the determinant:

\[ \Delta = 2\left(-36 - \frac{\lambda^2}{4}\right) + 5\left(15 - \frac{5\lambda}{4}\right) + \frac{5}{2}\left(-\frac{5\lambda}{2} - 30\right) \]

### Step 4: Simplify the expression
Expanding this gives:

\[ \Delta = -72 - \frac{\lambda^2}{2} + 75 - \frac{25\lambda}{4} - \frac{25\lambda}{4} - \frac{75}{2} \]

Combining like terms:

\[ \Delta = -\frac{\lambda^2}{2} - \frac{50\lambda}{4} + 75 - 72 - \frac{75}{2} \]

This simplifies to:

\[ \Delta = -\frac{\lambda^2}{2} - \frac{25\lambda}{2} + \frac{150 - 144}{2} = -\frac{\lambda^2 + 25\lambda + 6}{2} \]

Setting \( \Delta = 0 \):

\[ \lambda^2 + 25\lambda + 144 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula:

\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-25 \pm \sqrt{625 - 576}}{2} \]

This simplifies to:

\[ \lambda = \frac{-25 \pm 7}{2} \]

Calculating the two possible values:

1. \( \lambda = \frac{-25 + 7}{2} = \frac{-18}{2} = -9 \)
2. \( \lambda = \frac{-25 - 7}{2} = \frac{-32}{2} = -16 \)

### Step 6: Check the condition \( |\lambda| < 16 \)
We need to check which of these values satisfies \( |\lambda| < 16 \):

- For \( \lambda = -9 \), \( |-9| = 9 < 16 \) (valid)
- For \( \lambda = -16 \), \( |-16| = 16 \) (not valid)

Thus, the value of \( \lambda \) that satisfies the condition is:

\[ \lambda = -9 \]

### Final Answer:
The value of \( \lambda \) such that \( |\lambda| < 16 \) and the equation represents a pair of straight lines is:

\[ \lambda = -9 \]