Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The distance between the parallel lines represented by <img alt="\mathrm{ x^2+6 x y+9 y^2+4 x+12 y-5=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20x%5E2&plus

The distance between the parallel lines represented by

\mathrm{ x^2+6 x y+9 y^2+4 x+12 y-5=0}, is
 

Option: 1

\frac{1}{\sqrt{10}}
 


Option: 2

\frac{5}{\sqrt{10}}
 


Option: 3

\frac{6}{\sqrt{10}}

 


Option: 4

\frac{7}{\sqrt{10}}


Answers (1)

best_answer

The given equation can be written as

\mathrm{(x+3 y)^2+4(x+3 y)-5=0}

Or \mathrm{(x+3 y+5)(x+3 y-1)=0}

\therefore the lines are parallel and can be written in perpendicular form as

\mathrm{\frac{x+3 y}{\sqrt{1+9}}=\frac{-5}{\sqrt{1+9}} \text { and } \frac{x+3 y}{\sqrt{1+9}}=\frac{1}{\sqrt{1+9}} }

Required distance \mathrm{=\left|\frac{1}{\sqrt{10}}+\frac{5}{\sqrt{10}}\right|=\frac{6}{\sqrt{10}} . }

Hence option 3 is correct.

 

Posted by

Gaurav

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question