Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Let \mathrm{P \equiv(-1,0), Q \equiv(0,0)} and \mathrm{R \equiv(3,3 \sqrt{3})} be three points. Then the equation of the bisector of angle \mathrm{P Q R} is

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

In the figure observe that \mathrm{\tan \alpha=\frac{3 \sqrt{3}}{3}=\sqrt{3}}


\mathrm{\Rightarrow \quad \alpha=60^{\circ} \quad \angle P Q R=120^{\circ}}
Hence, the bisector makes an angle of \mathrm{\left(60^{\circ}+\alpha\right)}
i.e., \mathrm{60^{\circ}+60^{\circ}=120^{\circ}} with positive direction of \mathrm{x}-axis.

Hence its equation is \mathrm{y-0=\left(\tan 120^{\circ}\right)(x-0)}.

\mathrm{\therefore (c) }

Posted by

sudhir kumar

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question