Let $A(3-i), B(2+i)$ be two points in the argand plane. If the point $P$ represents the complex number…

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
  1. the circle with AB as diameter
  2. the line passing through A and B
  3. the perpendicular bisector of AB
  4. the ellipse with AB as major axis

Solution

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 $A B$

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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