Straight Lines
Triangle and Incentre
Grade 11
Question:
<p>The incentre of the triangle with vertices \((1,\sqrt{3})\), \((0,0)\) and \((2,0)\) is</p>
<p>\(\left(1,\dfrac{\sqrt{3}}{2}\right)\)</p>
<p>\(\left(\dfrac{2}{3},\dfrac{1}{\sqrt{3}}\right)\)</p>
<p>\(\left(\dfrac{2}{3},\dfrac{\sqrt{3}}{2}\right)\)</p>
<p>\(\left(1,\dfrac{1}{\sqrt{3}}\right)\)</p>
Step-by-Step Solution
Key Concept: The incentre divides the triangle based on side lengths as weights. Use the formula: I = (aA + bB + cC)/(a+b+c), where a,b,c are opposite side lengths to vertices A,B,C.
<p><strong>Step 1:</strong> Identify vertices: A = (1,√3), B = (0,0), C = (2,0)</p><p><strong>Step 2:</strong> Calculate side lengths (opposite to each vertex):</p><p>• a = |BC| = √[(2-0)² + (0-0)²] = 2</p><p>• b = |AC| = √[(2-1)² + (0-√3)²] = √(1+3) = 2</p><p>• c = |AB| = √[(1-0)² + (√3-0)²] = √(1+3) = 2</p><p><strong>Step 3:</strong> Triangle is equilateral (all sides equal). For equilateral triangle, incentre = centroid.</p><p><strong>Step 4:</strong> Apply incentre formula:</p><p>I = (a·A + b·B + c·C)/(a+b+c)</p><p>I = [2(1,√3) + 2(0,0) + 2(2,0)]/(2+2+2)</p><p>I = [(2,2√3) + (0,0) + (4,0)]/6</p><p>I = (6, 2√3)/6</p><p>∴ Answer: I = (1, √3/3)</p>
Correct Answer: D