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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.

Respuesta :

Answer:

The answer is "[tex]\sqrt{12.5}[/tex]"

Step-by-step explanation:

[tex]\to (3,2), \ (6,1)[/tex]

Distance[tex]=\sqrt{9+1}=\sqrt{10}[/tex]

midpoint height [tex]=\sqrt{10+2.5}=\sqrt{12.5}[/tex]

let

Gradient [tex]= \frac{1}{-3}[/tex]      

Perpendicular Gradient[tex]= 3[/tex]

midpoints [tex]= (4.5, 1.5)[/tex]

Similar to perpendicular

Now I'll obtain the perpendicular bisector solution

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