Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The distance of the line from the point <img alt="\mathrm{(1,1)}" src="https://

The distance of the line \mathrm{2 x-3 y=4} from the point \mathrm{(1,1)} in the direction of the line \mathrm{x+y=1} is

Option: 1

\sqrt{2}


Option: 2

5 \sqrt{2}


Option: 3

1 / \sqrt{2}


Option: 4

 none of these


Answers (1)

best_answer

Parametric equation of line \mathrm{PQ}

\mathrm{\frac{x-1}{-\frac{1}{\sqrt{2}}}=\frac{y-1}{\frac{1}{\sqrt{2}}}=r}
Any point on this line \mathrm{\left(1-\frac{r}{\sqrt{2}}, 1+\frac{r}{\sqrt{2}}\right)}
\mathrm{Now \, 2\left(1-\frac{r}{\sqrt{2}}\right)-3\left(1+\frac{r}{\sqrt{2}}\right)=4}

Hence the distance is \mathrm{\sqrt{2}}
Hence (A) is the correct answer.

Posted by

qnaprep

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 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question