Straight Lines
Equilateral triangle on a line
Grade 11
Question:
<p>The straight line \(3x + 4y - 12 = 0\) meets the coordinates axes at \(A\) and \(B\). An equilateral triangle \(ABC\) is constructed. The possible coordinates of vertex \(C\) are</p>
<p>\(\left(2\left(1 - \dfrac{3\sqrt{3}}{4}\right),\, \dfrac{3}{2}\left(1 - \dfrac{4}{\sqrt{3}}\right)\right)\)</p>
<p>\(\left(-2\left(1 + \sqrt{3}\right),\, 3/2\left(1 - \sqrt{3}\right)\right)\)</p>
<p>\(\left(2\left(1 + \sqrt{3}\right),\, 3/2\left(1 + \sqrt{3}\right)\right)\)</p>
<p>\(\left(2\left(1 + \dfrac{3\sqrt{3}}{4}\right),\, \dfrac{3}{2}\left(1 + \dfrac{4}{\sqrt{3}}\right)\right)\)</p>
Step-by-Step Solution
Key Concept: Find intercepts A and B on the axes, then use the distance AB and equilateral triangle geometry to construct point C at distance AB from both A and B on the perpendicular bisector of AB.
<p><strong>Step 1:</strong> Find intercepts of line 3x + 4y - 12 = 0</p><p>At x-axis (y = 0): 3x = 12 → x = 4, so A = (4, 0)</p><p>At y-axis (x = 0): 4y = 12 → y = 3, so B = (0, 3)</p><p><strong>Step 2:</strong> Calculate distance AB</p><p>AB = √[(4-0)² + (0-3)²] = √(16 + 9) = √25 = 5</p><p><strong>Step 3:</strong> Find midpoint M of AB</p><p>M = ((4+0)/2, (0+3)/2) = (2, 1.5)</p><p><strong>Step 4:</strong> Find perpendicular bisector direction</p><p>Slope of AB = (3-0)/(0-4) = -3/4</p><p>Slope of perpendicular bisector = 4/3</p><p>Direction vector along perpendicular: (3, 4) (normalized: unit vector)</p><p><strong>Step 5:</strong> Calculate height of equilateral triangle</p><p>Height h = (√3/2) × 5 = (5√3)/2</p><p><strong>Step 6:</strong> Find unit vector perpendicular to AB</p><p>Perpendicular direction to AB (3, -4): unit vector = ±(4, 3)/5</p><p><strong>Step 7:</strong> Calculate point C</p><p>C = M ± h × (unit perpendicular vector)</p><p>C = (2, 1.5) ± (5√3/2) × (4/5, 3/5)</p><p>C = (2, 1.5) ± (2√3, 1.5√3)</p><p><strong>C₁ = (2 + 2√3, 1.5 + 1.5√3) and C₂ = (2 - 2√3, 1.5 - 1.5√3)</strong></p><p>∴ Answer: A</p>
Correct Answer: A