How does one go about solving $arg(\frac{z-2i}{z-6}) = \frac{1}{2}\pi$ This should give $$\frac{z-2i}{z-6} = bi$$
but solving that gives me $$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$
and substituting $z$ for $x + yi$ gives me $x = \frac{-2b +6b^2}{1+b^2}$ and $y=\frac{-6b +2}{1+b^2}$
And I have no clue how to continue now. The answer is suppose to be $x = -by$, but I have no clue how to get there (have I made a calculation error??)
EDIT: Thanks everyone!!

As you said solving gets you here:
$$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$
and finally, substituting gets u here:
$$x = \frac{-2b +6b^2}{1+b^2}$$ and $$y=\frac{-6b +2}{1+b^2}$$
Just divide above two equations you get
$$x = \frac{-2b + 6b^2}{1 + b^2} = -\frac{2 - 6b}{1 + b^2}\cdot b = -by$$
$$\implies x+by=0$$
