<p>If one vertex of an equilateral triangle of side 'a' lies at origin and the other lies on the line $x - \sqrt{3}y = 0$, then the coordinates of the third vertex are:</p>
<p>(a) $(0, a)$</p>
<p>(b) $\left(\frac{\sqrt{3}a}{2}, -\frac{a}{2}\right)$</p>
<p>(c) $(0, -a)$</p>
<p>(d) $\left(-\frac{\sqrt{3}a}{2}, \frac{a}{2}\right)$</p>
Step-by-Step Solution
Key Concept: Use the constraint that two vertices are at the origin and on the line x - √3y = 0, both at distance 'a' apart, then find the third vertex at distance 'a' from both using rotation properties of equilateral triangles.
<p><strong>Step 1: Set up the constraints.</strong> Let the three vertices be A (at origin), B (on the line), and C (unknown). We have A = (0, 0) and need |AB| = |AC| = |BC| = a.</p><p><strong>Step 2: Find vertex B on the line.</strong> Since B lies on x - √3y = 0, we have B = (√3t, t) for some parameter t. The distance from origin is |AB| = √(3t² + t²) = 2|t| = a, so t = ±a/2.</p><p><strong>Step 3: Consider both cases for B.</strong> Taking t = a/2 gives B = (√3a/2, a/2). (We'll verify this leads to the answer.)</p><p><strong>Step 4: Use rotation to find C.</strong> The third vertex C of an equilateral triangle is obtained by rotating B about A by ±60°. Using the rotation matrix for -60° (clockwise):</p><p>C = (√3a/2 · cos(-60°) - a/2 · sin(-60°), √3a/2 · sin(-60°) + a/2 · cos(-60°))</p><p>C = (√3a/2 · 1/2 + a/2 · √3/2, -√3a/2 · √3/2 + a/2 · 1/2)</p><p>C = (√3a/4 + √3a/4, -3a/4 + a/4) = (√3a/2, -a/2)</p><p><strong>Step 5: Verify.</strong> Check |AC| = √((√3a/2)² + (a/2)²) = √(3a²/4 + a²/4) = √(a²) = a ✓</p><p>Check |BC| with B = (√3a/2, a/2): |BC| = √(0 + a²) = a ✓</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B