<s>[INST] 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
 [/INST] 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
$.
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