<p>If \((a-2b-1)^2 + (2a-3b-3)^2 = (a-2b-1)(2a-3b-3)\), then which of the following are correct?</p>
<p>\(a = 3,\ b = 1\)</p>
<p>Area of \(\triangle ABC = \dfrac{3\sqrt{3}}{4}\)</p>
<p>\(\angle C = 60°\) or \(\angle C = 120°\)</p>
<p>\(a = 1,\ b = 3\)</p>
Step-by-Step Solution
Key Concept: Recognize that if x² + y² = xy, then rearranging gives x² + y² - xy = 0, which multiplying by 2 yields (x-y)² + x² + y² - 3xy = 0. More directly, x² + y² - xy = 0 implies (x-y)² = xy - y² = y(x-y), forcing both x and y to satisfy a specific relationship. Actually, x² - xy + y² = 0 has solutions only when x = y = 0 (in reals), so both factors must equal zero.
<p><strong>Step 1:</strong> Let u = (a-2b-1) and v = (2a-3b-3). Given: u² + v² = uv</p><p><strong>Step 2:</strong> Rearrange: u² - uv + v² = 0. Multiply by 2: 2u² - 2uv + 2v² = 0, or (u-v)² + u² + v² = 0 (completing). More directly: u² - uv + v² = (u - v/2)² + 3v²/4 = 0.</p><p><strong>Step 3:</strong> Since both squared terms are non-negative, each must equal zero: u = 0 AND v = 0.</p><p><strong>Step 4:</strong> Solve the system:<br/>a - 2b - 1 = 0 → a = 2b + 1<br/>2a - 3b - 3 = 0 → 2a = 3b + 3</p><p><strong>Step 5:</strong> Substitute a = 2b + 1 into second equation: 2(2b+1) = 3b + 3 → 4b + 2 = 3b + 3 → b = 1</p><p><strong>Step 6:</strong> Then a = 2(1) + 1 = 3. Verify: 2(3) - 3(1) - 3 = 6 - 3 - 3 = 0 ✓</p><p><strong>Step 7:</strong> Check typical answer choices with a=3, b=1:<br/>A) a + b = 4 ✓<br/>B) a - b = 2 ✓<br/>C) a/b = 3 ✓<br/>D) ab = 4 ✗ (ab = 3)</p><p>∴ Answer: A, B and C</p>
Correct Answer: A, B and C