Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Lines <img alt="\mathrm{L_1 \equiv a x+b y+c=0 \text { and } L_2=1 x+m y+n=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BL_1%20%5Cequiv%20a%20x&plus;b%20y&plus;c%3D0%

Lines \mathrm{L_1 \equiv a x+b y+c=0 \text { and } L_2=1 x+m y+n=0} intersect at the point P and make an angle \theta with each  other. The equation of  a  line different  from  \mathrm{L_{2}} which passes through  P and  makes the same  angle \theta with \mathrm{L_{1}} is \mathrm{2(al + bm) (ax + by+ c) - k(lx + my + n) = 0} where \mathrm{k=}

Option: 1

\mathrm{a^{2}-b^{2}}


Option: 2

\mathrm{a^{2}-2b^{2}}


Option: 3

\mathrm{a^{2}+b^{2}}


Option: 4

\mathrm{-a^{2}-b^{2}}


Answers (1)

best_answer

Equation of PA is \mathrm{a x+b y+c=0}\: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(1)

and equation of PB is \mathrm{1 x+m y+n=0}\: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(2)

Now,  equation  of  any  line  through  the  point of  intersection of lines (1) and (2) is

\mathrm{\mathrm{l} x+m y+n+k(a x+b y+c)=0}\: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(3)

Slope of line (3) \mathrm{=-\frac{1+a k}{m+b k} \text {, }}

Slope of \mathrm{P B=-\frac{1}{m}} and slope of \mathrm{P A=-\frac{a}{b}}

Since, angle between \mathrm{\mathrm{PA} \text { and } \mathrm{PB}=\theta}

\mathrm{\Rightarrow \tan \theta=\left|\frac{-\frac{a}{b}+\frac{l}{m}}{1+\frac{a l}{b m}}\right|=\left|\frac{b l-a m}{b m+a l}\right|}\: \: \: \: \: \: \: \: \: \: \: \: \: .......(4)

Again if equation (3) represents line PC, then

\mathrm{\tan \theta=\left|\frac{-\frac{a}{b}+\frac{l+a k}{m+b k}}{1+\frac{1+a k}{m+b k} \cdot \frac{a}{b}}\right|=\left|\frac{l b-a m}{\left(a^2+b^2\right) k+l a+m b}\right|}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ......(5)

from (4) and (5), 

\mathrm{\frac{b l-a m}{b m+a l}= \pm \frac{l b-a m}{\left(a^2+b^2\right) k+l a+m b}}

\mathrm{\Rightarrow\left(a^2+b^2\right) k+l a+m b= \pm(b m+a l) \Rightarrow k=0 \text { or } \frac{-2(a l+b m)}{a^2+b^2}}

But from (3), k = 0 gives line PB, therefore equation of the line PC is

\begin{aligned} & \mathrm{ lx+m y+n-\frac{2(a l+b m)}{a^2+b^2}(a x+b y+c)=0} \\ \\& \Rightarrow \mathrm{2(a \mid+b m)(a x+b y+c)-\left(a^2+b^2\right)(\mid x+m y+n)=0} \end{aligned}

Posted by

himanshu.meshram

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 Mains 2026 Exam City Slip LIVE: NTA session 2 test centre allotment soon; admit card date
anam-khan

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 city slip soon; admit card, exam dates, here's how to prepare

Latest Question