#Abs(z-4+2i)=Abs(z+8-6i)#. In the form #asqrt13, ainQQ#, what is the least value of #absz#?

I think the cartesian form of the locus of #z# is #y=3/2x+5#, but I'm not sure. Thanks

1 Answer
Feb 16, 2018

#a=10/13#

Explanation:

#z# is a point that is equidistant from #(4-2i)# and #(-8+6i)# (since #|z-z_0|# is the distance between #z# and #z_0#)- that means in the equivalent #RR^2# plane it is the perpendicular bisector of the points #(4,-2)# and #(-8,6)#. The answer we are seeking is the shortest distance from the origin to this line.

Now the line satisfies the equation

#(x-4)^2+(y+2)^2 = (x+8)^2+(y-6)^2#

which simplifies to the equation

#3x-2y-10=0#

The distance of this line from the origin is

# {|3times 0-2 times 0-10|}/sqrt{3^2+2^2}=10/sqrt{13}=10/13 sqrt{13}#