If |z-2|=|z-1|,where z is a complex number,th | vedclass

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(B) Given $|z-2|=|z-1|$.Squaring both sides,we get $|z-2|^2 = |z-1|^2$.Using the property $|w|^2 = w \bar{w}$,we have $(z-2)(\bar{z}-2) = (z-1)(\bar{z

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Parallel to $y$-axis

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(B) Given $|z-2|=|z-1|$.Squaring both sides,we get $|z-2|^2 = |z-1|^2$.Using the property $|w|^2 = w \bar{w}$,we have $(z-2)(\bar{z}-2) = (z-1)(\bar{z}-1)$.Expanding both sides: $z\bar{z} - 2z - 2\bar{z} + 4 = z\bar{z} - z - \bar{z} + 1$.Simplifying the equation: $-2z - 2\bar{z} + 4 = -z - \bar{z} + 1$.Rearranging terms: $z + \bar{z} = 3$.Substituting $z = x + iy$ and $\bar{z} = x - iy$:$(x + iy) + (x - iy) = 3$.$2x = 3 \Rightarrow x = \frac{3}{2}$.This represents a vertical line passing through $x = 1.5$,which is parallel to the $y$-axis.

If $|z-2|=|z-1|$,where $z$ is a complex number,then the locus of $z$ is a straight line:

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