Complex Numbers And Quadratic Equations question 368
Question: If $ z=x+iy $ and $ |z-2+i|=|z-3-i|, $ then locus of z is [RPET 1999]
Options:
A) $ 2x+4y-5=0 $
B) $ 2x-4y-5=0 $
C) $ x+2y=0 $
D) $ x-2y+5=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ |z-2+i|=|z-3-i| $
$ \Rightarrow $ $ |(x-2)+i(y+1)| $ $ =|(x-3)+i(y-1)| $
$ \Rightarrow $ $ \sqrt{{{(x-2)}^{2}}+{{(y+1)}^{2}}} $ = $ \sqrt{{{(x-3)}^{2}}+{{(y-1)}^{2}}} $
$ \Rightarrow $ $ x^{2}+4-4x+y^{2}+1+2y=x^{2}+9-6x+y^{2}+1-2y $
$ \Rightarrow $ $ 2x+4y-5=0 $ .
BETA
