It is about vectors.. But nobody could explain it..?

Consider a complex number (z=x+yj) in the "complex plane" to be a vector. The vertices of a triangle A, B, and C are given by the complex numbers 1+j, 2-j, and -1 respectively. Find the point that is equidistant to A, B, and C.

It is about vectors.. But nobody could explain it..?
z=x+yj is (x,y) and the points (1,1) , (2,-1) and (-1,0) respectively.





Let the distance be called D.





D^2 = (x -1)^2 + (y - 1)^2 ..(1)





D^2 = (x -2)^2 + (y +1)^2 ....(2)





D^2 = (x +1)^2 + y ^2 ....(3)





3 equations and 3 unknowns... Solve it simulteneously and you are done.
Reply:Here's a shot in the dark from a rusty math major.





In the complex plane, the distance between two points is equal the the magnitude of the difference of those two points.





With that information, you can set up an general point X that is equidistant from all of the vertices.





You would end up with something like |A-X|=|B-X|=|C-X|





This should allow you to set up a system of linear equations with three equations and three unknowns, the distanct from the points, the real component of the point, and the imaginary component of the point.





Let me know if that works.