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JEE Main 2016Medium

Q.The foot of the perpendicular from (0,0)(0,0) to the line ax+by+c=0ax + by + c = 0 is

a
(−aca2+b2,−bca2+b2)\left( \dfrac{-ac}{a^2 + b^2}, \dfrac{-bc}{a^2 + b^2} \right)
b
(aca2+b2,bca2+b2)\left( \dfrac{ac}{a^2 + b^2}, \dfrac{bc}{a^2 + b^2} \right)
c
(−aca2+b2,−bca2+b2)\left( \dfrac{-a c}{a^2 + b^2}, \dfrac{-b c}{a^2 + b^2} \right)
d
(aca2+b2,bca2+b2)\left( \dfrac{a c}{a^2 + b^2}, \dfrac{b c}{a^2 + b^2} \right)

Correct Answer: Option C

The correct solution involves applying the fundamental concept to derive the final value step by step...

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