|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Jun 2008 20:52:06 IST
|
|
|
1 A point moves such that the sum of its distances from the four sides of a square of length ‘g’ is constant. Then what is the locus of the point??
|
|
|
|
10 Jun 2008 20:53:32 IST
|
|
|
it can only be the center of square..
|
IMPOSSIBLES ARE OFTEN UNTRIED... |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
10 Jun 2008 20:55:35 IST
|
|
|
NO DUDE IT CAN ALSO BE THE VERTICES OF THE SQUARE ...AM WORKING ON THE PROB. WILL POST IT IN A MIN
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
11 Jun 2008 00:26:19 IST
|
|
|
locus is points lying on and inside that square.
rate me if i m right.
correct me if i m wrong.
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
11 Jun 2008 00:32:33 IST
|
|
|
vikasbansal is correct and the sum of the distances will always be 2g (sum of length and breadth)
|
---------------------------------------------------------------
- Gaurav Ragtah (spideyunlimited)
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
11 Jun 2008 17:27:47 IST
|
|
|
Yes vikas is correct. For any point in or on the square the sum of distances will be 2g
|
Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
2
