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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Jul 2007 19:24:11 IST
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Find Locus of point P(x,y) such that sum of its distances from two fixed perpendicular straight lines is 2 units.
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God has given you one face, and you
make yourself another.
~William Shakespeare
You were born an original. Don't die a copy.
~John Mason |
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Let the first line be ax + by = c
The line perpendicular to this line will be : bx - ay = c
Let pt be (h,k) (to avoid confusion)
dist from first line = (ah + bk - c)/(a^2 + b^2)^1/2
dist from sec line = (bh - ak - c)/(a^2 + b^2)^1/2
form the eqn, u will get the answer
rater if this helps u...
cheers!!!
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26 Jul 2007 19:33:26 IST
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sankydreams has given the best possible answer according to me......good one yaar.............
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The inevitable truth of life.....everyone in our life is going 2 hurt sooner or later......u just have 2 realise who is worth.....
the PAIN or the PERSON...!!! |
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26 Jul 2007 19:38:51 IST
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yup
good answer
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I always like to walk in rain as no one can see me crying there :(
frnds are like diamonds , if u hit them , they don't break but they slip frm ur hands
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4 Aug 2007 13:56:55 IST
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best answer would be
Let the 2 perpendicular lines be X axis and Y axis
thus eqn will be
| x | + | y | = 2 ...(distance from x axis = x coordinate, and from y axis = y coordinate)
put values of x and y as 2, 1 , 0 and see the points of intersection with x and y axes
locus is a tilted square with coordinates as (2,0) , (0,2)
, (-2,0) and (0, -2)
PLZ RATE :)
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---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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