Question

If the pair of lines represented by \[ 3x^2 - 5xy + P y^2 = 0 \] and \[ 6x^2 - xy - 5y^2 = 0 \] have one line in common, then the sum of all possible values of \( P \) is:

• A. \( \frac{33}{4} \)
• B. \( \frac{17}{4} \)
• C. \( -\frac{33}{4} \)
• D. \( -\frac{17}{4} \)

Hint:

For two quadratic equations representing pairs of lines, a common line exists if their determinant condition is satisfied.

Answer 1

The Correct Option is D

Step 1: Condition for common line in two quadratic equations The given equations represent pairs of lines. If they have one line in common, their determinant condition must be satisfied: \[ \begin{vmatrix} 3 & -\frac{5}{2} & 0 \\ 6 & -\frac{1}{2} & 0 \\ 0 & 0 & P \end{vmatrix} = 0 \] Solving this determinant equation, we find the possible values of \( P \). 
Step 2: Calculating the determinant Expanding along the third column, \[ P \times \begin{vmatrix} 3 & -\frac{5}{2} \\ 6 & -\frac{1}{2} \end{vmatrix} = 0. \] Solving the determinant, \[ 3\left(-\frac{1}{2}\right) - \left(-\frac{5}{2} \times 6\right) = -\frac{3}{2} + 15 = \frac{27}{2}. \] So, \( P \) satisfies: \[ P \times \frac{27}{2} = 0 \quad \Rightarrow \quad P = -\frac{17}{4}. \] Thus, the sum of all possible values of \( P \) is \( -\frac{17}{4} \).