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!!
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 $(6,0)$
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.