## Task B-2.6.

The equations $|z-2 a|=|z-a| \text{ and } |z-3 a i|=|z-a i|, a \in \mathbb{R}, a \neq 0, z \in \mathbb{C}$, determine sets of points in the Gaussian plane that, together with the coordinate axes, enclose a figure with an area of 75 square units. Calculate the perimeter of this figure.
## Solution.

If $z=x+y i, x, y \in \mathbb{R}$, the equation $|z-2 a|=|z-a|$ transforms into

$$
\begin{aligned}
& \sqrt{(x-2 a)^{2}+y^{2}}=\sqrt{(x-a)^{2}+y^{2}} \\
& (x-2 a)^{2}+y^{2}=(x-a)^{2}+y^{2} \\
& x^{2}-4 a x+4 a^{2}=x^{2}-2 a x+a^{2} \\
& 2 a x=3 a^{2} \\
& x=\frac{3}{2} a
\end{aligned}
$$

Thus, the first set of points is a line perpendicular to the $x$-axis, $x=\frac{3}{2} a$.

By an analogous procedure, the second equation $|z-3 a i|=|z-a i|$ transforms into

$$
\sqrt{x^{2}+(y-3 a)^{2}}=\sqrt{x^{2}+(y-a)^{2}}
$$

which, after simplification, gives $y=2 a$, a line perpendicular to the $y$-axis.

The obtained lines, together with the coordinate axes, form a rectangle with side lengths $\frac{3}{2}|a|$ and $2|a|$.

It follows that $P=\frac{3}{2} \cdot 2 a^{2}=75 \Rightarrow a=5$ or $a=-5$.

The perimeter of the obtained figure is $O=2 \cdot \frac{3}{2} a+2 \cdot 2 a=7 a=35$ units.

Note: If a student recognizes that the equations $|z-2 a|=|z-a|$ and $|z-3 a i|=|z-a i|$ represent the perpendicular bisectors of segments, this should be recognized and all the points for that part of the task should be awarded.