Question

There exist two complex numbers $c$, say $c_1$ and $c_2$, so that $3 + 2i$, $6 + i$, and $c$ form the vertices of an equilateral triangle. Find the product $c_1 c_2$ in rectangular form.

Answer 1

Answer:

The answer is "$\sqrt{12.5}$"

Step-by-step explanation:

$\to (3,2), \ (6,1)$

Distance$=\sqrt{9+1}=\sqrt{10}$

midpoint height $=\sqrt{10+2.5}=\sqrt{12.5}$

let

Gradient $= \frac{1}{-3}$      

Perpendicular Gradient$= 3$

midpoints $= (4.5, 1.5)$

Similar to perpendicular

Now I'll obtain the perpendicular bisector solution

An equation of the center circle$(4.5,1.5)$ radius will be$\sqrt{12.5}$.