Question

If $|z_2+i{z}_1|=|z_1|+|z_2|$ and $|z_1|=3$ and $|z_2|=4$, then area of $△ABC$, if affixes of $A,B$ and $C$ are $z_1,z_2$ and $\left[\frac{z_2-i{z}_1}{1-i}\right]$ respectively, is,

• A. $\frac{5}{2}$ sq. units
• B. $0$ sq. unit
• C. $\frac{25}{2}$ sq. units
• D. $\frac{25}{4}$ sq. units

Answer 1

The correct option is D $\frac{25}{4}$ sq. units

$|z_2+i{z}_1|=|z_1|+|z_2|\Rightarrow |z_2+i{z}_1|=|z_2|+|i{z}_1|$
$\Rightarrow i{z}_1,0+i0$ and $z_2$ are collinear
$\Rightarrow \arg \left(i{z}_1\right)=\arg \left(z_2\right)$
$\Rightarrow \arg \left(z_2\right)-\arg \left(z_1\right)=\frac{\pi }{2}$

Let,
$z_3=\frac{z_2-i{z}_1}{1-i}$
$\Rightarrow (1-i)z_3=z_2-i{z}_1$
$\Rightarrow z_2-z_3=i(z_1-z_3)$
$\therefore \angle ACB=\frac{\pi }{2}$ and
$|z_1-z_3|=|z_2-z_3|$
$\Rightarrow AC=BC$
$\because A{B}^2=A{C}^2+B{C}^2$
$\Rightarrow AC=\frac{5}{\sqrt{2}}(\because AB=5)$
Therefore, area of $△ABC$
$=\left(\frac{1}{2}\right)AC\times BC=\frac{25}{4}$ sq. units