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.

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 !!!!!!!!!!)

SIR WHAT ABOUT THIS APPROACH???

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.