Question

`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

Answer 1

`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}`