Quantitative Aptitude Ques 1633
Question: Locus of the points equidistant from the points $ (-1,2) $ and (3, 4) is
Options:
A) $ 2x+y-5=0 $
B) $ 2x+y+10=0 $
C) $ 2x-y-10=0 $
D) $ 2x-y+10=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Let $ A,(-1,2) $ and $ B,(3,4) $ and $ p,(x-y) $ be such that $ PA^{2}=PB^{2} $
$ \Rightarrow $ $ {{(x+1)}^{2}}+{{(y-2)}^{2}}={{(x-3)}^{2}}+{{(y-4)}^{2}} $
$ \Rightarrow $ $ x^{2}+1+2x+y^{2}+4-4y=x^{2}+9-6x $ $ +,y^{2}+16-8y $
$ \Rightarrow $ $ 2x+5-4y=-,6x-8y+25 $
$ \Rightarrow $ $ 8x+4y-20=0 $
$ \Rightarrow $ $ 2x+y-5=0 $
BETA
