<p><b>Paragraph for Question nos. 622 and 623</b><br>Consider the following set of points in the \(x\)-\(y\) plane \(A = \{(a, b) \mid a, b \in I \text{ and } |a| + |b| \leq 2\}\).</p><p>The number of triangles whose vertices are points in \(A\), is:</p>
Step-by-Step Solution
Key Concept: First identify all lattice points satisfying |a| + |b| ≤ 2 (forms a diamond), then count total possible triangles C(n,3) and subtract collinear triples to get valid triangles.
<p><strong>Step 1: Find all points in set A where |a| + |b| ≤ 2</strong></p><p>Points: (0,0), (±1,0), (0,±1), (±2,0), (0,±2), (±1,±1)</p><p>Total: 9 points (diamond pattern centered at origin)</p><p><strong>Step 2: Count total combinations</strong></p><p>Total ways to choose 3 points from 9: C(9,3) = 84</p><p><strong>Step 3: Identify and count collinear triples</strong></p><p>Horizontal lines: {(-2,0), (0,0), (2,0)} → 1 triple; {(-1,1), (0,1), (1,1)} → 1 triple; {(-1,-1), (0,-1), (1,-1)} → 1 triple</p><p>Vertical lines: {(0,-2), (0,0), (0,2)} → 1 triple</p><p>Diagonal lines (slope 1): {(-1,-1), (0,0), (1,1)} → 1 triple</p><p>Diagonal lines (slope -1): {(-1,1), (0,0), (1,-1)} → 1 triple</p><p>Total collinear triples: 6</p><p><strong>Step 4: Calculate valid triangles</strong></p><p>Number of triangles = 84 - 6 = 78</p><p>∴ Answer: C</p>
Correct Answer: C