Equation of locus of points satisfied by $\frac{\left|z+3i\right|}{\left|z-6i\right|}=1$ Equation of locus of points satisfied by $\frac{\left|z+3i\right|}{\left|z-6i\right|}=1$
The answer I got is $y=\frac{3}{2}$, but the answer given in my book is $y=-0.5x+2.25$
Can anyone please confirm which is the right answer
Edit: Sorry, the modulus was on each numerator and denominator, not for the whole thing, though I don't think this makes a difference.
My working
subbing $z=x+yi$ gives
$\frac{\left|x+yi+3i\right|}{\left|x+yi-6i\right|}=1$
$\frac{x^2+(y+3)^2}{x^2+(y-6)^2}=1^2$
Expanding brackets and then solving gives
$y=3/2$
Thank You
 A: To cut down on unanswered questions, here we go!
The book's answer is certainly wrong, as one readily sees by considering $z=\frac94i.$
Now, we clearly cannot have $z=6i$ as a solution, for then we have $\frac90=1,$ which is nonsensical. Consequently, the given equation is equivalent to $$|z+3i|=|z-6i|,$$ or, put another way, to $$\bigl|z-(-3i)\bigr|=|z-6i|.$$
Since $|z-w|$ is the distance from $z$ to $w$ for all $z,w\in\Bbb C,$ then the equation above says that $z$ is equidistant from $-3i$ and $6i.$ Readily, putting $z=x+iy,$ this is equivalent to saying that $(x,y)$ is equidistant from $(0,-3)$ and $(0,6),$ i.e.: $$\sqrt{x^2+(y+3)^2}=\sqrt{x^2+(y-6)^2}.$$ This is, of course, equivalent to your approach, and solving the preceding equation yields $y=\frac32,$ as you say. 
A: Given two points,
the set of points
equidistant from them
is the perpendicular bisector
of the line joining them.
If the points are
$(a, b)$ and
$(c,d)$,
the midpoint is
$((a+c)/2, (b+d)/2)$.
The slope of the line is
$\dfrac{d-b}{c-a}$,
so the slope of the normal
is the negative reciprocal
of this or
$-\dfrac{c-a}{d-b}
=\dfrac{a-c}{d-b}
$.
Therefore the equation
of the perpendicular bisector is
$\dfrac{y-(b+d)/2)}{x-(a+c)/2}
=\dfrac{a-c}{d-b}
$.
If $a=c=0$,
as in this case,
the equation is
$\dfrac{y-(b+d)/2)}{x}
=0
$
or
$y=(b+d)/2$.
In this case,
$b=-3$ and $d=6$,
so the equation is
$y
=(6-3)/2
=3/2
$.
