<p>The area of the region represented by \(|x - y| \leq 2\) and \(|x + y| \leq 2\) is:</p>
Step-by-Step Solution
Key Concept: The region defined by |x - y| ≤ 2 and |x + y| ≤ 2 forms a square when the two inequalities are viewed as boundaries of the form y = x ± 2 and y = -x ± 2. Use coordinate transformation or recognize this as a rotated square with diagonals along the coordinate axes.
<p><strong>Step 1:</strong> Expand the inequalities:<br>|x - y| ≤ 2 means -2 ≤ x - y ≤ 2, giving lines x - y = 2 and x - y = -2<br>|x + y| ≤ 2 means -2 ≤ x + y ≤ 2, giving lines x + y = 2 and x + y = -2</p><p><strong>Step 2:</strong> Find the four vertices by solving pairs of boundary lines:<br>• x - y = 2 and x + y = 2: Adding gives 2x = 4, so (2, 0)<br>• x - y = 2 and x + y = -2: Adding gives 2x = 0, so (0, -2)<br>• x - y = -2 and x + y = 2: Adding gives 2x = 0, so (0, 2)<br>• x - y = -2 and x + y = -2: Adding gives 2x = -4, so (-2, 0)</p><p><strong>Step 3:</strong> The region is a square with vertices at (2, 0), (0, 2), (-2, 0), (0, -2). The diagonals have length 4 each and are perpendicular (along the x and y axes).</p><p><strong>Step 4:</strong> For a rhombus/square with perpendicular diagonals of lengths d₁ = 4 and d₂ = 4:<br>Area = ½ × d₁ × d₂ = ½ × 4 × 4 = 8</p><p>∴ Answer: A (Area = 8)</p>
Correct Answer: A