Question

If $|z_{2} + iz_{1}|= |z_{1}| + |z_{2}|$ and $|z_{1}| = 3$ and $|z_{2}| = 4,$ then the area of $\Delta ABC,$ if vertices of $A, B,$ and C are $z_{1}, z_{2},$ and $[(z_{2} -iz_{1})/ (1-i)]$ respectively, is _____.

Answer 1

$|z_{2}+iz_{1}|=|z_{1}|+|z_{2}|$
$\Rightarrow iz_{1}, 0+i0$ and $z_{2}$ are collinear
$\Rightarrow $ arg $(iz_{1})=$ arg$(z_{2})$
$\Rightarrow $ arg $(z_{2})-$ arg $(z_{1})=\frac{\pi}{2}$
Let $z_{3}=\frac{ z_{2}-iz_{1}}{1-i}$
or $\left(1-i\right) z_{3} =z_{2} -iz_{1}$
or $z_{2} -z_{3} =i\left(z_{1}-z_{3}\right)$
$\therefore ∠ ACB =\frac{\pi}{2}$
and $\left|z_{1}-z_{3}\right|=\left|z_{2}-z_{3}\right|$
$\Rightarrow AC=BC$
$\because AB^{2} =AC^{2}+BC^{2}$
$\Rightarrow AC =\frac{5}{\sqrt{2}} \left(\because AB=5\right)$
Therefore, area of $\Delta ABC$ is $(1/2)AC \times BC = AC^{2}/2 = 25/4$ sq. units.