The point P denotes the complex number z=x+iy | vedclass

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(B) Let $z = x + iy$. Then $\frac{2z-i}{z-2} = \frac{2(x+iy)-i}{(x+iy)-2} = \frac{2x + i(2y-1)}{(x-2) + iy}$.To make this purely real,multiply the num

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$x+4y-2=0$ and $(x, y) 
eq(2,0)$

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(B) Let $z = x + iy$. Then $\frac{2z-i}{z-2} = \frac{2(x+iy)-i}{(x+iy)-2} = \frac{2x + i(2y-1)}{(x-2) + iy}$.To make this purely real,multiply the numerator and denominator by the conjugate of the denominator,$(x-2) - iy$:$\frac{[2x + i(2y-1)][(x-2) - iy]}{(x-2)^2 + y^2} = \frac{2x(x-2) + y(2y-1) + i[(2y-1)(x-2) - 2xy]}{(x-2)^2 + y^2}$.For the expression to be purely real,the imaginary part must be zero:$(2y-1)(x-2) - 2xy = 0$.$2xy - 4y - x + 2 - 2xy = 0$.$-x - 4y + 2 = 0$,which simplifies to $x + 4y - 2 = 0$.Since the denominator $z-2 
eq 0$,we must have $(x, y) 
eq (2, 0)$.

The point $P$ denotes the complex number $z=x+iy$ in the Argand plane. If $\frac{2z-i}{z-2}$ is a purely real number,then the equation of the locus of $P$ is

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