<p>The <em>x</em>-coordinate of the incentre of the triangle that has the coordinates of midpoints of its sided as \((0, 1)\), \((1, 1)\) and \((1, 0)\) is</p>
Step-by-Step Solution
Key Concept: The midpoints of a triangle's sides form the medial triangle. To find the original triangle's vertices, use the property that if M₁, M₂, M₃ are midpoints of sides BC, CA, AB respectively, then the original vertices can be recovered using: A = M₃ + M₂ - M₁ (and cyclic permutations). The incentre's x-coordinate is the weighted average of vertex x-coordinates using opposite side lengths as weights.
<p><strong>Step 1:</strong> Let the vertices be A, B, C. Given midpoints: M₁ = (0,1) [midpoint of BC], M₂ = (1,1) [midpoint of CA], M₃ = (1,0) [midpoint of AB].</p><p><strong>Step 2:</strong> Using the relationship between vertices and midpoints:<br>A = M₂ + M₃ - M₁ = (1,1) + (1,0) - (0,1) = (2,0)<br>B = M₃ + M₁ - M₂ = (1,0) + (0,1) - (1,1) = (0,0)<br>C = M₁ + M₂ - M₃ = (0,1) + (1,1) - (1,0) = (0,2)</p><p><strong>Step 3:</strong> Calculate side lengths:<br>a = BC = √[(0-0)² + (2-0)²] = 2<br>b = CA = √[(2-0)² + (0-2)²] = 2√2<br>c = AB = √[(0-2)² + (0-0)²] = 2</p><p><strong>Step 4:</strong> Incentre formula: I = (aA + bB + cC)/(a+b+c)<br>x-coordinate = (2·2 + 2√2·0 + 2·0)/(2 + 2√2 + 2) = 4/(4 + 2√2) = 4/[2(2 + √2)] = 2/(2 + √2)<br>Rationalize: = 2(2 - √2)/[(2 + √2)(2 - √2)] = 2(2 - √2)/(4 - 2) = 2(2 - √2)/2 = 2 - √2</p><p>∴ Answer: A (which is 2 - √2)</p>
Correct Answer: A