Quantitative Aptitude Ques 1748

Question: If a point $ (x,y) $ in $ xy $ -plane is equidistant from $ (-1,1) $ and $ (4,3) $ then

Options:

A) $ 10x+4y=23 $

B) $ 6x+4y=23 $

C) $ -x+y=7 $

D) $ 4x+3y=0 $

Show Answer

Answer:

Correct Answer: A

Solution:

  • Distance between $ (x,y) $ and $ (-1,1) $ $ =\sqrt{{{(y-1)}^{2}}+{{(x+1)}^{2}}} $ Distance between $ (x,y) $ and $ (4,3) $ $ =\sqrt{{{(y-3)}^{2}}+{{(x-4)}^{2}}} $ $ \because $ Point are equidistant $ =\sqrt{{{(y-1)}^{2}}+{{(x+1)}^{2}}}=\sqrt{{{(y-3)}^{2}}+{{(x-4)}^{2}}} $ On squaring both side, we get
    $ {{(y-1)}^{2}}+{{(x+1)}^{2}}={{(y-3)}^{2}}+{{(x-4)}^{2}} $

$ \Rightarrow $ $ y^{2}+1-2y+x^{2}+1+2x $ $ =y^{2}+9-6y+x^{2}+16-8x $

$ \Rightarrow $ $ 2x-2y+2=-6y-8x+25 $

$ \therefore $ $ 10x+4y=23 $

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