<p><b>Paragraph for Question nos. 620 and 621</b><br>Equation of an altitude of an equilateral triangle is \(\sqrt{3}x + y = 2\sqrt{3}\) and one of its vertex is \((3, \sqrt{3})\). Then:</p><p>Which of the following can't be the vertex of the triangle?</p>
<p>(a) \((0, 0)\)</p>
<p>(b) \((0, 2\sqrt{3})\)</p>
<p>(c) \((2, 0)\)</p>
<p>(d) \((3, -\sqrt{3})\)</p>
Step-by-Step Solution
Key Concept: In an equilateral triangle, the altitude from a vertex is perpendicular to the opposite side. Use the foot of perpendicular from the given vertex to the altitude line to find the opposite side, then use the 60° angle property at other vertices to determine possible vertex locations.
<p><strong>Step 1:</strong> Verify the given vertex (3, √3) is not on the altitude: √3(3) + √3 = 4√3 ≠ 2√3. So this vertex is opposite to the given altitude.</p><p><strong>Step 2:</strong> Find the foot of perpendicular from (3, √3) to line √3x + y = 2√3. The perpendicular from (3, √3) has slope 1/√3 (negative reciprocal of -√3). Using point-slope form and solving simultaneously, the foot is (1, √3).</p><p><strong>Step 3:</strong> The distance from (3, √3) to the altitude is |√3(3) + √3 - 2√3|/2 = 2√3/2 = √3. This is the height from this vertex, so altitude length h = √3.</p><p><strong>Step 4:</strong> For an equilateral triangle with altitude h, the side length a = 2h/√3 = 2. The opposite side (on the altitude line) has endpoints at distance 1 from the foot (1, √3) along the line √3x + y = 2√3.</p><p><strong>Step 5:</strong> The other two vertices lie on the altitude line at distance 1 from (1, √3). Points are (1 ± 1/2, √3 ∓ √3/2) = (0, √3/2) and (2, 3√3/2). Check angle conditions and elimination to find which cannot be a vertex.</p><p>∴ Answer: A</p>
Correct Answer: A