Question:

<p><strong>Step 1: Identify the vertices and calculate side vectors</strong></p><p>Let A = z₁, B = z₂, C = (1-i)z₁ + iz₂</p><p>Side vector AB = z₂ - z₁</p><p>Side vector AC = (1-i)z₁ + iz₂ - z₁ = -iz₁ + iz₂ = i(z₂ - z₁)</p><p>Side vector BC = (1-i)z₁ + iz₂ - z₂ = (1-i)z₁ + (i-1)z₂ = (1-i)(z₁ - z₂)</p></p><p><strong>Step 2: Check if the triangle is isosceles by comparing side lengths</strong></p><p>|AB| = |z₂ - z₁|</p><p>|AC| = |i(z₂ - z₁)| = |i|·|z₂ - z₁| = |z₂ - z₁|</p><p>Since |AB| = |AC|, the triangle is isosceles with AB = AC.</p></p><p><strong>Step 3: Check if the triangle is right-angled</strong></p><p>For right angle at A, we need AB ⊥ AC. Two vectors are perpendicular if one is a purely imaginary multiple of the other.</p><p>Notice that AC = i(z₂ - z₁) = i·AB</p><p>This means AC is obtained by rotating AB by 90° counterclockwise (multiplication by i).</p><p>Therefore, AB ⊥ AC, and the triangle has a right angle at A.</p></p><p><strong>Step 4: Verify using the perpendicularity condition</strong></p><p>For perpendicularity: (AB)·(AC̄) should be purely imaginary</p><p>(z₂ - z₁)·[i(z₂ - z₁)]̄ = (z₂ - z₁)·(-i)(z̄₂ - z̄₁) = -i|z₂ - z₁|²</p><p>This is purely imaginary (real part = 0), confirming perpendicularity.</p></p><p><strong>∴ Answer: C</strong> The triangle ABC is both isosceles (AB = AC) and right-angled (∠BAC = 90°).</p>