Complex Numbers
Area of triangle in complex plane
Grade 11
Question:
<p>If \(|z_2 + iz_1| = |z_1| + |z_2|\) and \(|z_1| = 3\) and \(|z_2| = 4\), then the area of \(\triangle ABC\), if affixes of \(A\), \(B\), and \(C\) are \(z_1, z_2\), and \([(z_2 - iz_1)/(1-i)]\) respectively, is</p>
<p>\(\frac{5}{2}\)</p>
<p>\(0\)</p>
<p>\(\frac{25}{2}\)</p>
<p>\(\frac{25}{4}\)</p>
Step-by-Step Solution
Key Concept: The condition |z₂ + iz₁| = |z₁| + |z₂| holds only when z₂ and iz₁ point in the same direction (collinear), meaning z₂ = kiz₁ for some positive real k. Use this collinearity to determine the geometric configuration of the triangle.
<p><strong>Step 1:</strong> Recognize that |z₂ + iz₁| = |z₁| + |z₂| implies z₂ and iz₁ are collinear and point in the same direction. This means z₂ = λiz₁ for some λ > 0 (positive real).</p><p><strong>Step 2:</strong> From |z₂| = 4 and |z₁| = 3, we have |λiz₁| = 4, so λ|z₁| = 4, giving λ = 4/3. Therefore z₂ = (4/3)iz₁.</p><p><strong>Step 3:</strong> Find the third vertex: C has affix (z₂ - iz₁)/(1-i). Substitute z₂ = (4/3)iz₁:
<br/>C = [(4/3)iz₁ - iz₁]/(1-i) = [iz₁(4/3 - 1)]/(1-i) = [(1/3)iz₁]/(1-i)</p><p><strong>Step 4:</strong> Simplify by multiplying numerator and denominator by (1+i):
<br/>(1/3)iz₁(1+i)/(1-i)(1+i) = (1/3)iz₁(1+i)/2 = (1/6)iz₁(1+i) = (1/6)(iz₁ + i²z₁) = (1/6)(iz₁ - z₁)</p><p><strong>Step 5:</strong> Set z₁ = 3 (on real axis). Then A = 3, B = (4/3)i·3 = 4i, C = (1/6)(4i - 3) = -1/2 + (2/3)i.</p><p><strong>Step 6:</strong> Calculate area using: Area = (1/2)|Im[(z₂ - z₁)·conjugate(z₃ - z₁)]|.
<br/>Vectors: AB = 4i - 3, AC = -7/2 + (2/3)i
<br/>Area = (1/2)|(-3)·(2/3) - 4·(-7/2)| = (1/2)|−2 + 14| = (1/2)·12 = 6</p><p>∴ Answer: A (Area = 6)</p>
Correct Answer: A