Straight Lines
Triangle centres
Grade 11
Question:
<p>Let \(A \equiv (1, \sqrt{3})\), \(B \equiv (0, 0)\), \(C \equiv (2, 0)\). The incentre of \(\triangle ABC\) is:</p>
<p>(A) \(\left(1, \dfrac{1}{\sqrt{3}}\right)\)</p>
<p>(B) \(\left(\dfrac{2}{3}, \dfrac{1}{\sqrt{3}}\right)\)</p>
<p>(C) \(\left(\dfrac{2}{3}, \sqrt{3}\right)\)</p>
<p>(D) \(\left(1, \sqrt{3}\right)\)</p>
Step-by-Step Solution
Key Concept: The incenter divides the triangle in the ratio of the sides. Use the formula: Incenter = (aA + bB + cC)/(a+b+c), where a, b, c are the lengths of sides opposite to vertices A, B, C respectively.
<p><strong>Step 1:</strong> Find the side lengths of triangle ABC.</p><p>A = (1, √3), B = (0, 0), C = (2, 0)</p><p>Side BC (opposite to A): a = |BC| = 2</p><p>Side AC (opposite to B): b = |AC| = √[(2-1)² + (0-√3)²] = √[1 + 3] = 2</p><p>Side AB (opposite to C): c = |AB| = √[(1-0)² + (√3-0)²] = √[1 + 3] = 2</p><p><strong>Step 2:</strong> The triangle is equilateral with all sides = 2.</p><p><strong>Step 3:</strong> Apply the incenter formula: I = (aA + bB + cC)/(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>I = (1, √3/3)</p><p>∴ Answer: A</p>
Correct Answer: A