Straight Lines
Angle between lines
Grade 11
Question:
<p>A straight line <i>L</i> through the point (3, −2) is inclined at an angle of 60° to the line \(\sqrt{3}x + y = 1\). If <i>L</i> also intersects the <i>x</i>-axis, then the equation of <i>L</i> is</p>
<p>\(y + \sqrt{3}x + 2 - 3\sqrt{3} = 0\)</p>
<p>\(y - \sqrt{3}x + 2 + 3\sqrt{3} = 0\)</p>
<p>\(\sqrt{3}y - x + 3 + 2\sqrt{3} = 0\)</p>
<p>\(\sqrt{3}y + x - 3 + 2\sqrt{3} = 0\)</p>
Step-by-Step Solution
Key Concept: Use the angle between two lines formula: tan(θ) = |(m₁ - m₂)/(1 + m₁m₂)|. First find the slope of the given line, then apply this formula with θ = 60° to find the slope of line L, ensuring the line passes through (3, -2) and intersects the x-axis.
<p><strong>Step 1:</strong> Find the slope of the given line √3x + y = 1, or y = -√3x + 1. Therefore, m₂ = -√3.</p><p><strong>Step 2:</strong> Use the angle formula: tan(60°) = |(m₁ - (-√3))/(1 + m₁(-√3))| = |(m₁ + √3)/(1 - √3m₁)|</p><p><strong>Step 3:</strong> Since tan(60°) = √3, we have: √3 = |(m₁ + √3)/(1 - √3m₁)|</p><p><strong>Step 4:</strong> This gives two cases:</p><p>Case 1: √3(1 - √3m₁) = m₁ + √3 → √3 - 3m₁ = m₁ + √3 → m₁ = 0</p><p>Case 2: √3(1 - √3m₁) = -(m₁ + √3) → √3 - 3m₁ = -m₁ - √3 → m₁ = √3</p><p><strong>Step 5:</strong> For m₁ = 0: Line is y = -2 (horizontal). This doesn't intersect x-axis (except at infinity).</p><p><strong>Step 6:</strong> For m₁ = √3: Using point-slope form with point (3, -2): y + 2 = √3(x - 3) → y = √3x - 3√3 - 2 → √3x - y - 3√3 - 2 = 0</p><p><strong>Step 7:</strong> Verify: When y = 0: √3x - 3√3 - 2 = 0 → x = 3 + (2/√3) (intersects x-axis ✓)</p><p>∴ Answer: A</p>
Correct Answer: A