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If a point $P$ denotes the complex number $z=x+$ iy in the Argand plane and if $\frac{z-(2+i)}{z+(1-2 i)}$ is purely real, then the locus of $\mathrm{P}$ is
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Verified Answer
The correct answer is:
the line $x+3 y-5=0$ excluding the point $(-1,2)$
$\frac{Z-(2+i)}{Z+(1-2 i)}$
$Z$ is purely real
$$
\begin{aligned}
& \therefore \quad \frac{(x+i y)-2-i}{x+i y+1-2 i}=\frac{(x-2)+(y-1) i}{(x+1)+(y-2) i} \\
& =\frac{[(x-2)+(y-1) i][(x+1)-(y-2) i]}{[(x+1)+(y-2 i)][(x+1)-(y-2) i]} \\
& \{(x-2)(x+1)+(y-1)(y-2)\} \\
& =\frac{+i\{(x+1)(y-1)-(x-2)(y-2)\}}{(x+1)^2+(y-2)^2} \\
& \text { Imaginary part }=0 \\
& \therefore \quad \frac{(x+1)(y-1)-(x-2)(y-2)}{(x+1)^2+(y-2)^2}=0 \\
& \therefore \quad(x+1)(y-1)-(x-2)(y-2)=0 \text { and } x \neq-1, y \neq 2 \\
& \Rightarrow x y-x+y-1-x y+2 x+2 y-4=0 \\
& \Rightarrow x+3 y-5=0 \text { and } x \neq-1, y \neq 2 \\
&
\end{aligned}
$$
$Z$ is purely real
$$
\begin{aligned}
& \therefore \quad \frac{(x+i y)-2-i}{x+i y+1-2 i}=\frac{(x-2)+(y-1) i}{(x+1)+(y-2) i} \\
& =\frac{[(x-2)+(y-1) i][(x+1)-(y-2) i]}{[(x+1)+(y-2 i)][(x+1)-(y-2) i]} \\
& \{(x-2)(x+1)+(y-1)(y-2)\} \\
& =\frac{+i\{(x+1)(y-1)-(x-2)(y-2)\}}{(x+1)^2+(y-2)^2} \\
& \text { Imaginary part }=0 \\
& \therefore \quad \frac{(x+1)(y-1)-(x-2)(y-2)}{(x+1)^2+(y-2)^2}=0 \\
& \therefore \quad(x+1)(y-1)-(x-2)(y-2)=0 \text { and } x \neq-1, y \neq 2 \\
& \Rightarrow x y-x+y-1-x y+2 x+2 y-4=0 \\
& \Rightarrow x+3 y-5=0 \text { and } x \neq-1, y \neq 2 \\
&
\end{aligned}
$$
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