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!!
 A: The expression means that z subtends an angle $\frac{\pi}{2}$ at the points $2i$ and $6$
Ponder upon the following visual

The green arrows show some more positions which are possible for $z$. Clearly, as it subtends $90^o$ at those points, so the locus of $z$ is a semicircle, the end point of hose diameter is $(0,2)$ and $(6,0)$
A: Most of the time, polar form is better when dealing with complex numbers. Draw a circle with $2i$ and $6$ as the ends of its diameter. This diameter divide the circle into 2 parts, the lower half is the loci.
Hint: if $AB$ is the diameter of a circle and $P$ is on the circle, $\angle APB =\pm \frac{\pi}{2}$
A: You have not made a mistake. Observe that $x$ is a multiple of $y$:
$$x = \frac{-2b + 6b^2}{1 + b^2} = -b \cdot \frac{2 - 6b}{1 + b^2} = -by$$
which is what the answer states.
A: 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
$$x = \frac{-2b + 6b^2}{1 + b^2} = -\frac{2 - 6b}{1 + b^2}\cdot b = -by$$
$$\implies x+by=0$$
