Question

If the triangle formed by the complex coordinates $A\left(z\right),B(2+3i),C(4+5i)$ which satisfy the relations $|z-(2+3i)|=|z-(4+5i)|$ and $|z-(3+4i\left)\right|\le 4$, then

• A. ar. $(△ABC{)}_{max}=4\sqrt{2}$ sq. unit
• B. ar. $(△ABC{)}_{max}=4$ sq. unit
• C. if area of $△ABC$ is maximum, then unequal angle is $<\frac{\pi }{4}$
• D. if area of $△ABC$ is maximum, then unequal angle is $\ge \frac{\pi }{4}$

Answer 1

Answer: (A), (C)

if area of $△ABC$ is maximum, then unequal angle is $<\frac{\pi }{4}$


As given $|z-(2+3i)|=|z-(4+5i)|$
$\Rightarrow AB=AC$
Hence $△ABC$ will be isosceles triangle
and lenght of perpendicular from $A$ on $BC$ will be meet at mid point of $BC$
$D=(z_1+z_2)/2=3+4i$
Hence length of perpendicular
$=|z-(3+4i\left)\right|=4$ (for maximum value)
So required maximum area
$=\frac{1}{2}\times 4\times |z_1-z_2|=4\sqrt{2}$ sq. unit
Now let
$\frac{2+3i-z}{4+5i-z}=e^{i\theta }\Rightarrow \tan \theta /2=\frac{DC}{AD}=\frac{1}{2\sqrt{2}}\Rightarrow \tan \theta =\frac{4\sqrt{2}}{7}\therefore \arg \left(\frac{2+3i-z}{4+5i-z}\right)=\theta <\frac{\pi }{4}$