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$|2z-1| = |z-2|$:

Rearrange to $|z-\frac12| = \frac12|z-2|$: distance from $\frac12$ is half distance from $2$.

So it's a circle (distance from one point is a constant multiple $-$ not $0$ or $1$ $-$ of distance from another point). You know two points on the diameter: $1$ and $-1$. So you know the centre and radius.

$|z-8|+|z+8|=20$:

Distance from one point + distance from another point is constant: an ellipse with the two points as foci.

$|z-2| = |z-3i|$:

Distance from one point = distance from another point: this is the perpendicular bisector of the two points.

$|2z+3| = |z+6|$:

Rearrange to $|z+\frac32| = \frac12|z+6|$ and it's the same case as the first equation.