Complex Numbers And Quadratic Equations question 369
Question: Locus of the point z satisfying the equation $ |iz-1| $ + $ |z-i|=2 $ is [Roorkee 1999]
Options:
A) A straight line
B) A circle
C) An ellipse
D) A pair of straight lines
Show Answer
Answer:
Correct Answer: A
Solution:
$ |iz-1|+|z-i|=2 $
$ \Rightarrow $ $ |i(x+iy)-1|+|x+iy-i|=2 $
$ \Rightarrow $ $ |-(y+1)+ix|+|x+i(y-1)|=2 $
$ \Rightarrow $ $ \sqrt{{{(-(y+1))}^{2}}+x^{2}}+\sqrt{x^{2}+{{(y-1)}^{2}}}=2 $
$ \Rightarrow $ $ \sqrt{{{(y+1)}^{2}}+x^{2}}=2-\sqrt{x^{2}+{{(y-1)}^{2}}} $
$ \Rightarrow $ $ y^{2}+1+2y+x^{2}=4+x^{2}+y^{2}+1-2y-4\sqrt{x^{2}+{{(y-1)}^{2}}} $
$ \Rightarrow $ $ 4y=4-4\sqrt{x^{2}+{{(y-1)}^{2}}} $
$ \Rightarrow $ $ y=1-\sqrt{x^{2}+{{(y-1)}^{2}}} $
$ \Rightarrow $ $ x^{2}+{{(y-1)}^{2}}={{(1-y)}^{2}} $
$ \Rightarrow $ $ x^{2}+y^{2}+1-2y=1+y^{2}-2y $
$ \Rightarrow $ $ x^{2}=0\Rightarrow x=0 $ i.e. equation of straight line.
BETA
