If z2+iz1=z1+z2 and z1=3 and z2=4, then the area of △ABC, if affixes of A, B, and C are z1, z2, and [(z2 - iz1)/(1 - i)] respectively, is

$|z_{2} + {iz}_{1}| = |z_{1}| + |z_{2}| = |{iz}_{1}| + |z_{2}|$

$\Rightarrow {iz}_{1}, 0 + i 0 and z_{2} are collinear$

$\begin{array}{l} \Rightarrow \arg ({iz}_{1}) = \arg (z_{2}) \\ \Rightarrow \arg (z_{2}) - \arg (z_{1}) = \frac{π}{2} \end{array}$

Let $z_{3} = \frac{z_{2} - {iz}_{1}}{1 - i}$

$\begin{array}{l} or (1 - i) z_{3} = z_{2} - {iz}_{1} \\ or z_{2} - z_{3} = i (z_{1} - z_{3}) \end{array}$

$∴ ∠ ACB = \frac{π}{2}$

and $|z_{1} - z_{3}| = |z_{2} - z_{3}|$

$\begin{array}{ll} \Rightarrow AC = BC \\ ∵ {AB}^{2} = {AC}^{2} + {BC}^{2} \\ \Rightarrow AC = \frac{5}{\sqrt{2}} (∵ AB = 5) \end{array}$

Therefore, area of $△ ABC is (1 / 2) AC \times BC = {AC}^{2} / 2 = 25 / 4 sq. units$

If $|z_{2} + {iz}_{1}| = |z_{1}| + |z_{2}| and |z_{1}| = 3$ and $|z_{2}| = 4$ , then the area of $△ ABC$ , if affixes of A, B, and C are z₁, z₂, and [(z₂ - iz₁)/(1 - i)] respectively, is