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  • Locus of a point that is equidistant from the lines <img alt="\mathrm{\mathrm{x}+\mathrm{y}-2 \sqrt{2}=0 \: and\: \mathrm{x}+\mathrm{y}-\sqrt{2}=0}" src="https://entrancecorner.oncodecogs.com/gif.l

Locus of a point that is equidistant from the lines \mathrm{\mathrm{x}+\mathrm{y}-2 \sqrt{2}=0 \: and\: \mathrm{x}+\mathrm{y}-\sqrt{2}=0} is
 

Option: 1

\mathrm{x}+\mathrm{y}-5 \sqrt{2}=0


 


Option: 2

\mathrm{x+y-3 \sqrt{2}=0}
 


Option: 3

\mathrm{2 \mathrm{x}+2 \mathrm{y}-3 \sqrt{2}=0}
 


Option: 4

\mathrm{2 x+2 y-5 \sqrt{2}=0}


Answers (1)

best_answer

For any point \mathrm{\mathrm{P}(\mathrm{x}, \mathrm{y})} that is equidistant from given line, we have

\mathrm{ x+y-\sqrt{2}=-(x+y-2 \sqrt{2}) }

\mathrm{ \Rightarrow \quad 2 x+2 y-3 \sqrt{2}=0 . }

Hence option 3 is correct.



 

Posted by

Ritika Harsh

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