<p>The straight line represented by \(x^2 + mxy - 2y^2 + 3y - 1 = 0\) meets at</p>
<p>(a) \((-1/3, 2/3)\)</p>
<p>(b) \((-1/3, -2/3)\)</p>
<p>(c) \((1/3, 2/3)\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: A second-degree equation represents a pair of straight lines if and only if its discriminant equals zero. Here, we use the condition that the equation can be factored as a product of two linear expressions, which requires the discriminant Δ = h² - ab = 0.
<p><strong>Step 1:</strong> Rewrite the equation in standard form: x² + mxy - 2y² + 3y - 1 = 0</p><p>Comparing with ax² + 2hxy + by² + 2gx + 2fy + c = 0:</p><ul><li>a = 1</li><li>2h = m, so h = m/2</li><li>b = -2</li><li>2g = 0, so g = 0</li><li>2f = 3, so f = 3/2</li><li>c = -1</li></ul><p><strong>Step 2:</strong> For a pair of straight lines, the condition is:</p><p>Δ = abc + 2fgh - af² - bg² - ch² = 0</p><p>Substituting values:</p><p>(1)(-2)(-1) + 2(3/2)(0)(m/2) - (1)(3/2)² - (-2)(0)² - (-1)(m/2)² = 0</p><p>2 + 0 - 9/4 - 0 + m²/4 = 0</p><p>2 - 9/4 + m²/4 = 0</p><p>8/4 - 9/4 + m²/4 = 0</p><p>-1/4 + m²/4 = 0</p><p>m² = 1</p><p>m = ±1</p><p><strong>Step 3:</strong> The lines represented meet at a point (the intersection point). For m = 1 or m = -1, we can verify the lines intersect at a unique point.</p><p>∴ Answer: A</p>
Correct Answer: A