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Let $A(3-i), B(2+i)$ be two points in the argand plane. If the point $P$ represents the complex number $z=x+i y$, which satisfies $|z-3+i|=|z-2-i|$, then the locus of the point $P$ is
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Verified Answer
The correct answer is:
the perpendicular bisector of AB
We have a complex number $z=x+i y$
$$
\begin{aligned}
& \text { and }|z-3+i|=|z-2-i| \\
& \Rightarrow \quad|x+i y-3+i|=|x+i y-2-i| \\
& \Rightarrow \quad(x-3)^2+(y+1)^2=(x-2)^2+(y-1)^2 \\
& \Rightarrow \quad x^3-6 x+9+y^2+2 y+1 \\
& =x^2-4 x+4+y^2-2 y+1
\end{aligned}
$$
So, it represent a line
Point $A(3,-1)$ and $B(2,1)$
So, mid-point of $A B=\left(\frac{5}{2}, 0\right)$
$$
m_1=\text { slope of } A B=\frac{1-(-1)}{2-3}=-2
$$
Point $\left(\frac{5}{2}, 0\right)$ satisfies the equation $-2 x+4 y+5=0$ and slope of line $=m_2=\frac{1}{2}$
Now, $m_1 m_2=-2 \times \frac{1}{2}=-1$
So, line $-2 x+4 y+5$ is perpendicular to $A B$. Hence, locus of point $p$ is the perpendicular bisector of $A B$.
$$
\begin{aligned}
& \text { and }|z-3+i|=|z-2-i| \\
& \Rightarrow \quad|x+i y-3+i|=|x+i y-2-i| \\
& \Rightarrow \quad(x-3)^2+(y+1)^2=(x-2)^2+(y-1)^2 \\
& \Rightarrow \quad x^3-6 x+9+y^2+2 y+1 \\
& =x^2-4 x+4+y^2-2 y+1
\end{aligned}
$$
So, it represent a line
Point $A(3,-1)$ and $B(2,1)$
So, mid-point of $A B=\left(\frac{5}{2}, 0\right)$
$$
m_1=\text { slope of } A B=\frac{1-(-1)}{2-3}=-2
$$
Point $\left(\frac{5}{2}, 0\right)$ satisfies the equation $-2 x+4 y+5=0$ and slope of line $=m_2=\frac{1}{2}$
Now, $m_1 m_2=-2 \times \frac{1}{2}=-1$
So, line $-2 x+4 y+5$ is perpendicular to $A B$. Hence, locus of point $p$ is the perpendicular bisector of $A B$.
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