doubt in perpendicular condition

hmm.. i understand...

i never had think of the proof of m1m2 as -1 ... goodd for that

thanks dude..

Maths 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.

hmm.. good comment abhishek :)

SIR 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 !!!!!!!!!!)

Warning: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.