Straight Line Question 37

Question: Let $ P(-1,0), $ $ Q(0,0) $ and $ R(3,3\sqrt{3}) $ be three points. Then the equation of the bisector of the angle PQR is [IIT Screening 2002]

Options:

A) $ \frac{\sqrt{3}}{2}x+y=0 $

B) $ x+\sqrt{3}y=0 $

C) $ \sqrt{3}x+y=0 $

D) $ x+\frac{\sqrt{3}}{2}y=0 $

Show Answer

Answer:

Correct Answer: C

Solution:

  • Slope of QR = $ \frac{3\sqrt{3}-0}{3-0}=\sqrt{3} $ i.e., $ \theta =60^{o} $ Clearly, $ \angle PQR=120^{o} $ OQ is the angle bisector of the angle, so line OQ makes 120o with the positive direction of x-axis. Therefore equation of the bisector of $ \angle PQR $ is $ y=\tan 120^{o}x $ or $ y=-\sqrt{3}x $ i.e., $ \sqrt{3}x+y=0 $ .



sathee Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on

Please select your preferred language
कृपया अपनी पसंदीदा भाषा चुनें