SATHEE: Straight Line Question 327

Straight Line Question 327

Question: Equation of the perpendicular bisector of the line segment joining the points (7, 4) and (-1, -2), is [AMU 1979]

Options:

A) $ 4x-3y=15 $

B) $ 3x+4y=15 $

C) $ 4x+3y=15 $

D) None of these

Show Answer

Answer:

Correct Answer: C

Solution:

Midpoint $ \equiv (3,1) $ Slope of perpendicular $ =\frac{-1}{\frac{-2-4}{-1-4}}=\frac{-4}{3} $ Therefore the required equation is $ 4x+3y=4(3)+3(1)=15 $ .

BETA

sathee@iitk.ac.in

Media coverage of SATHEE

Play Store

App Store

The Ministry of Education has launched the SATHEE initiative in association with IIT Kanpur to provide free guidance for competitive exams. SATHEE offers a range of resources, including reference video lectures, mock tests, and other resources to support your preparation. Please note that participation in the SATHEE program does not guarantee clearing any exam or admission to any institute.

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 "I understand".

Straight Line Question 327

Question: Equation of the perpendicular bisector of the line segment joining the points (7, 4) and (-1, -2), is [AMU 1979]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

Menu