Straight Lines
Angle Bisector
Grade 11
Question:
<p>Let \(P = (-1,\, 0)\), \(Q = (0,\, 0)\) and \(R = (3,\, 3\sqrt{3})\) be three points. The equation of the bisector of the angle <em>PQR</em> is</p>
<p>\(\sqrt{3}x + y = 0\)</p>
<p>\(x + \dfrac{\sqrt{3}}{2}\,y = 0\)</p>
<p>\(\dfrac{\sqrt{3}}{2}\,x + y = 0\)</p>
<p>\(x + \sqrt{3}\,y = 0\)</p>
Step-by-Step Solution
Key Concept: The angle bisector from Q passes through Q and makes equal angles with lines QP and QR. Find the slopes of QP and QR, then use the angle bisector formula: the bisector has slope equal to the average direction of the two lines (via tangent addition formula or direction vectors).
<p><strong>Step 1:</strong> Find slopes of the two lines from Q.</p><p>Line QP: slope m₁ = (0-0)/(-1-0) = 0 (horizontal line along negative x-axis)</p><p>Line QR: slope m₂ = (3√3-0)/(3-0) = √3, so angle with x-axis is θ₂ = 60°</p><p><strong>Step 2:</strong> The angle bisector makes equal angles with both lines.</p><p>Line QP makes angle 180° with positive x-axis. Line QR makes angle 60° with positive x-axis.</p><p>The angle between them is 120°. The bisector splits this equally, so it makes angles 180° - 60° = 120° and 60° from each line respectively.</p><p><strong>Step 3:</strong> The bisector direction is at angle (180° + 60°)/2 = 120° from positive x-axis, or equivalently at 60° above the positive x-axis going through Q.</p><p>Wait: Bisector angle = (0° + 60°)/2 = 30° from the positive x-axis (taking the interior angle).</p><p>Slope of angle bisector = tan(30°) = 1/√3 = √3/3</p><p><strong>Step 4:</strong> Equation passes through Q(0,0) with slope √3/3:</p><p>y - 0 = (√3/3)(x - 0)</p><p>y = (x/√3) or x - √3y = 0 or equivalently y = (√3/3)x</p><p>∴ Answer: A</p>
Correct Answer: A