wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If 4a2+9b2c2+12ab=0, then the family of straight lines ax+by+c=0 is concurrent at

A
(2,3) or (2,3)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(2,3) or (2,6)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(2,4) or (2,3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(2,5) or (1,5)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A (2,3) or (2,3)

We have,

4a2+9b2c2+12ab=0

Above equation can be written as

(2a+3b)2=c2

2a+3b=c or 2a+3b=c

2a+3bc=0 ...(1)

2a+3b+c=0 ...(2)

By comparing this to ax+by+c=0,

x=2,y=3 and x=2,y=3

Two points where the family of straight lines ax+by+c=0 is concurrent is (2,3) and (2,3).

Hence, option A is the correct answer.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon