Straight Lines
Pair of straight lines
Grade 11
Question:
<p>If the lines represented by the equation \(3y^2 - x^2 + 2\sqrt{3}x - 3 = 0\) are rotated about the point \((\sqrt{3}, 0)\) through an angle 15°, one clockwise direction and other in anti-clockwise direction, so that they become perpendicular, then the equation of the pair of lines in the new position is</p>
<p>(a) \(y^2 - x^2 + 2\sqrt{3} + 3 = 0\)</p>
<p>(b) \(y^2 - x^2 + 2\sqrt{3}x - 3 = 0\)</p>
<p>(c) \(y^2 - x^2 - 2\sqrt{3}x + 3 = 0\)</p>
<p>(d) \(y^2 - x^2 + 3 = 0\)</p>
Step-by-Step Solution
Key Concept: First, identify the pair of lines from the homogeneous equation by completing the square, then find their slopes. Rotate both lines by ±15° about the point (√3, 0) and use the condition that perpendicularity is maintained after rotation to find the new pair of lines.
<p><strong>Step 1:</strong> Rewrite the equation by completing the square: <br/>3y² - x² + 2√3x - 3 = 0<br/>3y² - (x² - 2√3x + 3) = 0<br/>3y² - (x - √3)² = 0<br/>(√3y - x + √3)(√3y + x - √3) = 0</p><p><strong>Step 2:</strong> The two lines are: x - √3y - √3 = 0 and x + √3y - √3 = 0, intersecting at (√3, 0). Their slopes are m₁ = 1/√3 and m₂ = -1/√3, making angles 30° and -30° with x-axis.</p><p><strong>Step 3:</strong> After rotating by ±15°, the lines make angles (30° + 15°) = 45° and (-30° - 15°) = -45° with the x-axis. These have slopes m'₁ = tan(45°) = 1 and m'₂ = tan(-45°) = -1.</p><p><strong>Step 4:</strong> The lines with slopes 1 and -1 passing through (√3, 0) are:<br/>y - 0 = 1(x - √3) ⟹ x - y - √3 = 0<br/>y - 0 = -1(x - √3) ⟹ x + y - √3 = 0</p><p><strong>Step 5:</strong> The combined equation is:<br/>(x - y - √3)(x + y - √3) = 0<br/>x² - y² - 2√3x + 3 = 0<br/>∴ <strong>Answer: x² - y² - 2√3x + 3 = 0</strong></p>
Correct Answer: D