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.

The answer is ""

Step-by-step explanation:

Distance

midpoint height

let

Gradient      

Perpendicular Gradient

midpoints

Similar to perpendicular

Now I'll obtain the perpendicular bisector solution

An equation of the center circle radius will be.

move it to the other side

step-by-step explanation:

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w