<p>Each side of a square is of length 4 units. The center of the square is at (3, 7) and one of the diagonals is parallel to the line \(y = x\). If the vertices of the square be (x₁, y₁), (x₂, y₂), (x₃, y₃) and (x₄, y₄) then find the value of \(\max(y_1, y_2, y_3, y_4) - \min(x_1, x_2, x_3, x_4)\).</p>
Step-by-Step Solution
Key Concept: When a square has its center at a given point and one diagonal parallel to y = x, the diagonals make 45° angles with the axes. Use the diagonal length (related to side length) and rotation to find vertex coordinates.
<p><strong>Step 1: Understand the setup.</strong> The square has side length 4, center at C(3, 7), and one diagonal is parallel to y = x (slope = 1).</p><p><strong>Step 2: Find the diagonal length.</strong> For a square with side a = 4, the diagonal length is d = 4√2. Each vertex is at distance d/2 = 2√2 from the center.</p><p><strong>Step 3: Determine vertex positions.</strong> If diagonal is parallel to y = x, it has direction vector (1, 1) (normalized: (1/√2, 1/√2)). The perpendicular diagonal has direction (-1, 1) (normalized: (-1/√2, 1/√2)).</p><p><strong>Step 4: Locate the four vertices.</strong> The vertices lie along both diagonals at distance 2√2 from center (3, 7):<br/>• Along diagonal y = x direction: (3, 7) ± 2√2·(1/√2, 1/√2) = (3, 7) ± (2, 2)<br/> → Vertices: (5, 9) and (1, 5)<br/>• Along perpendicular diagonal: (3, 7) ± 2√2·(-1/√2, 1/√2) = (3, 7) ± (-2, 2)<br/> → Vertices: (1, 9) and (5, 5)</p><p><strong>Step 5: Verify these form a square.</strong> Vertices: A(5, 9), B(1, 9), C(1, 5), D(5, 5)<br/>Side length: |AB| = 4 ✓, All sides equal 4 ✓, Center: ((5+1)/2, (9+5)/2) = (3, 7) ✓</p><p><strong>Step 6: Calculate max(y₁, y₂, y₃, y₄) and min(x₁, x₂, x₃, x₄).</strong><br/>y-coordinates: {9, 9, 5, 5}, so max = 9<br/>x-coordinates: {5, 1, 1, 5}, so min = 1<br/>Therefore: max(y) - min(x) = 9 - 1 = 8</p><p><strong>∴ Answer: 8</strong></p>
Correct Answer: 8