Quantitative Aptitude Ques 1997
Question: Locus of the points equidistant from the points $ (-1,-1) $ and (4, 2) is
Options:
A) $ 5x-3y-9=0 $
B) $ 5x+3y+9=0 $
C) $ 5x+3y-9=0 $
D) $ 5x-3y+9=0 $
E) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ A(-1,-1), $ $ B(4,2) $ and $ P(x,y) $ be the points such that $ AP^{2}=BP^{2} $
$ \therefore $ $ {{(x+1)}^{2}}+{{(y+1)}^{2}}={{(x-4)}^{2}}+{{(y-2)}^{2}} $ $ (x^{2}+1+2x+y^{2}+1+2y) $ $ =(x^{2}+16-8x+y^{2}+4-4y) $
$ \Rightarrow $ $ 2+2x+2y=20-8x-4y $
$ \Rightarrow $ $ 10x+6y-18=0 $
$ \Rightarrow $ $ 5x+3y-9=0 $
BETA
