1hmm.. good comment abhishek :)
m1.m2 =-1
is the condition for perpendicular lines.. how is x=0 and y=0 perpendicular?
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8 Answers
62hmm.. interesting...
i had to think :)
actually see the condition that is m1m2=-1 comes from another place:
Look closeley..
let the angle of first line from x axis is θ, 2nd line will have θ+Π/2
so slopes will be tanθ and tan(θ+Π/2)
tan(θ+Π/2) =cot{Π/2-(θ+Π/2)}
=-cotθ
thus, m1m2 = tanθ . -cotθ = -1 (but only when both tanθ and cotθ are defined....
So ur doubt is genuine... and the condition holds only when θâ‰
0 or Î /2... (nÎ /2) !!!!
So, technically (if one looks very very closely), in all such problems we should solve separately for these cases! (but we generally dont... and my best guess is that u wont be penalised for leaving this case!) and almost always u will get the right answer :)
I hope this makes it a bit clear?
1hmm.. i understand...
i never had think of the proof of m1m2 as -1 ... goodd for that
thanks dude..
33Maths doesnot capture ∞ but gives a hint of it. In this case m1=0 m2=∞ so m1m2 is indeterminate form so limit has 2 be calculated here also maths give us a hint.
11SIR JEE PERFECTLY ANALYSED
but mein iss mein thodi apni taang addana chahta hon :P
i was wondering y this could not be solved with the help of limits so let me try!!!!!
Let the slope of the line at the x axis be
Taking the R.H.L
Limit tan(0+h)
x--->0
and the line perpendicular be
Limit tan(pi/2+h)
h---->0
we know that m1 X m2 =-1
so
Limit tanh X( -coth)=-1 (∞ X 0)
h---->0
we can write this equation as
Limit tanh X (-1/tanh)=-1
h---->0
so we get
-1=-1
HENCE PROVED
SIMILARLY WE CAN CHECK ITS L.H.L AND WE WILL FIND THAT IT ALSO SATISFIES THE EQUATION
SO LIMIT EXISTS
HENCE PROVED
(CORRECT ME IF WRONG !!!!!!!!!!)
62see this approach is ok
but the bigger question is why do we not consider the case when m1=0 !!
that we should ideally do seperately!
24Warning:I could be wrong with the approach
Consider I quadrant as a pair of lines
=> Combined eqn of pair of lines :0.x2 +xy+0.y2=0
for pair of lines to be perpendicular:
coeff (x2)+coeff( y2)=0
which is true........
so x=0 and y=0 are perpendicular to each other.