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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

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    $\begingroup$ Please show us your workings, i.e. your work from which you "got is $y=\frac 32$. Other wise, we cannot confirm your solution. So, show us how you arrived at your answer, by editing your post to include it. $\endgroup$ – Namaste Aug 22 '17 at 20:11
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    $\begingroup$ I think you're right. The equation says point $z$ is equidistant from $-3i$ and $6i$, which should be a horizontal line at $y=1.5$. $\endgroup$ – Jirapat Samranvedhya Aug 22 '17 at 20:12
  • $\begingroup$ @amWhy I have done so now. Thanks $\endgroup$ – NumberCruncher Aug 22 '17 at 20:22
  • $\begingroup$ @JirapatSamranvedhya Thanks very much, that's how I thought about it first too. $\endgroup$ – NumberCruncher Aug 22 '17 at 20:22
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    $\begingroup$ NumberCruncher Thank you for including your work! $\endgroup$ – Namaste Aug 22 '17 at 22:10
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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.

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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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