Question

If $z=x+iy$ and $\left|z-2+i\right|=\left|z-3-i\right|$, then the locus of $z$ is

• A. $2x+4y-5=0$
• B. $2x-4y-5=0$
• C. $x+2y=0$
• D. $x+2y+5=0$

Answer 1

The correct option is A

$2x+4y-5=0$

Explanation for the correct option.

Find the locus of point $z$.

In the equation $\left|z-2+i\right|=\left|z-3-i\right|$ substitute $x+iy$ for $z$.

$$\begin{gathered} \left|x+iy-2+i\right|=\left|x+iy-3-i\right| \\ \Rightarrow \left|\left(x-2\right)+\left(y+1\right)i\right|=\left|\left(x-3\right)+\left(y-1\right)i\right| \\ \Rightarrow \sqrt{{\left(x-2\right)}^2+{\left(y+1\right)}^2}=\sqrt{{\left(x-3\right)}^2+{\left(y-1\right)}^2}\left[z=x+iy,\left|z\right|=\sqrt{x^2+y^2}\right] \end{gathered}$$

Now square both sides and form the equation.

$$\begin{gathered} {\left(\sqrt{{\left(x-2\right)}^2+{\left(y+1\right)}^2}\right)}^2={\left(\sqrt{{\left(x-3\right)}^2+{\left(y-1\right)}^2}\right)}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(y+1\right)}^2={\left(x-3\right)}^2+{\left(y-1\right)}^2 \\ \Rightarrow x^2-4x+4+y^2+2y+1=x^2-6x+9+y^2-2y+1 \\ \Rightarrow 2x+4y-5=0 \end{gathered}$$

So the locus of $z$ is $2x+4y-5=0$.

Hence, the correct option is A.