Straight Lines
Angle bisectors
Grade 11
Question:
<p>Let \(u \equiv ax + by + a\sqrt[3]{b} = 0\), \(v \equiv bx - ay + b\sqrt[3]{a} = 0\), \(a, b \in \mathbb{R}\) be two straight lines. The equations of the bisectors of the angle formed by \(k_1 u - k_2 v = 0\) and \(k_1 u + k_2 v = 0\) for non-zero real \(k_1\) and \(k_2\) are</p>
<p>(a) \(u = 0\)</p>
<p>(b) \(k_2 u + k_1 v = 0\)</p>
<p>(c) \(k_2 u - k_1 v = 0\)</p>
<p>(d) \(v = 0\)</p>
Step-by-Step Solution
Key Concept: The angle bisectors of two lines L₁ and L₂ are given by the equations obtained when we set the combined equations k₁L₁ - k₂L₂ = 0 and k₁L₁ + k₂L₂ = 0 equal to zero. The bisectors are simply u = 0 and v = 0 themselves, since the family of lines k₁u ± k₂v = 0 represents all lines passing through the intersection point.
<p><strong>Step 1:</strong> Note that k₁u - k₂v = 0 and k₁u + k₂v = 0 represent two distinct families of lines (for different values of k₁, k₂).</p><p><strong>Step 2:</strong> Adding these equations: (k₁u - k₂v) + (k₁u + k₂v) = 0 ⟹ 2k₁u = 0 ⟹ u = 0</p><p><strong>Step 3:</strong> Subtracting these equations: (k₁u - k₂v) - (k₁u + k₂v) = 0 ⟹ -2k₂v = 0 ⟹ v = 0</p><p><strong>Step 4:</strong> Verify that u = 0 and v = 0 are indeed perpendicular: The coefficients of u are (a, b) and of v are (b, -a). Dot product: a·b + b·(-a) = ab - ab = 0 ✓</p><p><strong>Step 5:</strong> Since u = 0 and v = 0 pass through the intersection point of k₁u - k₂v = 0 and k₁u + k₂v = 0, and are perpendicular to each other, they form the angle bisectors.</p><p><strong>∴ Answer:</strong> The bisectors are <strong>u = 0</strong> and <strong>v = 0</strong> (i.e., ax + by + a∛b = 0 and bx - ay + b∛a = 0)</p>
Correct Answer: A