<p>If one of the lines given by the equation \(2x^2 + pxy + 3y^2 = 0\) coincide with one of those given by \(2x^2 + qxy - 3y^2 = 0\) and the other lines represented by them be perpendicular, then</p>
Step-by-Step Solution
Key Concept: A pair of lines ax² + 2hxy + by² = 0 shares exactly one line with another pair if they have a common factor. Use the condition that one line is common and the other two are perpendicular (product of slopes = -1) to find p and q.
<p><strong>Step 1:</strong> Let the first pair 2x² + pxy + 3y² = 0 have lines l₁ and l₂, and the second pair 2x² + qxy - 3y² = 0 have lines l₂ and l₃ (where l₂ is common).</p><p><strong>Step 2:</strong> For pair 1: sum of slopes = -p/2, product = 3/2. For pair 2: sum of slopes = -q/2, product = -3/2.</p><p><strong>Step 3:</strong> Let slopes of l₁, l₂, l₃ be m₁, m₂, m₃. Then: m₁ + m₂ = -p/2, m₁m₂ = 3/2 and m₂ + m₃ = -q/2, m₂m₃ = -3/2.</p><p><strong>Step 4:</strong> Since l₁ ⊥ l₃: m₁m₃ = -1. From m₁m₂ = 3/2 and m₂m₃ = -3/2, we get: m₁m₃ = (m₁m₂)(m₂m₃)/m₂² = (3/2)(-3/2)/m₂² = -9/(4m₂²) = -1.</p><p><strong>Step 5:</strong> This gives m₂² = 9/4, so m₂ = ±3/2.</p><p><strong>Step 6:</strong> Testing m₂ = 3/2: From m₁m₂ = 3/2, we get m₁ = 1. From m₂m₃ = -3/2, we get m₃ = -1. Check: m₁m₃ = -1 ✓</p><p><strong>Step 7:</strong> From m₁ + m₂ = 1 + 3/2 = 5/2 = -p/2, we get p = -5. From m₂ + m₃ = 3/2 - 1 = 1/2 = -q/2, we get q = -1.</p><p>∴ Answer: B (p = -5, q = -1)</p>
Correct Answer: B