Question

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

• A. -10
• B. -9
• C. 10
• D. 9

Hint:

For $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ to be a pair of lines, $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.

Answer 1

The Correct Option is B

Step 1: Identify Coefficients
$a=2, h=-5, b=12, g=5/2, f=\lambda/2, c=-3$.
Step 2: Condition for Pair of Lines
The determinant $\Delta = 0$:
$\begin{vmatrix} a & h & g
h & b & f
g & f & c \end{vmatrix} = \begin{vmatrix} 2 & -5 & 5/2
-5 & 12 & \lambda/2
5/2 & \lambda/2 & -3 \end{vmatrix} = 0$.
Step 3: Expand and Solve
$2(-36 - \frac{\lambda^2}{4}) + 5(15 - \frac{5\lambda}{4}) + \frac{5}{2}(\frac{-5\lambda}{2} - 30) = 0$.
$-72 - \frac{\lambda^2}{2} + 75 - \frac{25\lambda}{4} - \frac{25\lambda}{4} - 75 = 0$.
$\lambda^2 + 25\lambda + 144 = 0$.
Step 4: Select Correct $\lambda$
$(\lambda + 9)(\lambda + 16) = 0 \Rightarrow \lambda = -9$ or $\lambda = -16$.
Given $|\lambda|<16$, the answer is $\lambda = -9$.
Final Answer: (b)