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Product Of Slopes Of X,y Axes=0

  1. May 6, 2005 #1
    This Is Strange !!

    Slope Of The Line X= 0[y- Axis] Is =+1
    Slope Of The Line Y=0 [x-axis] Is= 0
    Product Of The Slopes = 0

    Now Product Of Two Perpendicular Lines Should Be = - 1

    Is There A Contradiction ??
  2. jcsd
  3. May 6, 2005 #2


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    It's funny how both the y-axis and the line y=x have the same gradient according to you :tongue:
  4. May 8, 2005 #3
    well ,
    its even funnier how lack of reading skills can hamper a persons understanding .

    if u pay attention sire ,u'll find that i said
    that the line y=0 ,which you would agree is the x - axis and has a slope = 0
  5. May 8, 2005 #4


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    I don't disagree with you, but do you know what the gradient of the line x=0 (the y-axis) is?

    If you work that out you will understand where you went wrong. Please don't be so offended I was just lightly trying to point out your mistake.
  6. May 8, 2005 #5


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    This reminds me of the "proof" that 1 = 2. The mistake involved dividing by zero or something. ;)
  7. May 8, 2005 #6


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    Good advice. You should take it yourself. Zurtex did not say anything about "the line y=0, which you would agree is the x-axis". He specifically referred to the "y-axis" and his point was that its slope is not 1!
  8. May 8, 2005 #7

    James R

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    The slope of the y axis (x=0) is infinite. And, as we all know, infinity times zero equals negative 1. :)
  9. May 9, 2005 #8
    oopsey !!
    tan 45 syndrome !!!
    sorry for that ,

    but i thought anything multiplied by 0 is = 0
    infinity multiplied by zero being -1 is brand new information to me ,can anyone tell me how this so ??
  10. May 9, 2005 #9


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    Infinity multiplied by 0 is not -1, heck it doesn't even make sense. But what you can do is work out the limit of the 2 sides when multiplied by each other, which goes something like this:

    [tex]\lim_{x \rightarrow \infty} -x \frac{1}{x} = -1[/tex]

    Now although it's true that:

    [tex]\lim_{x \rightarrow \infty} -x = -\infty[/tex]


    [tex]\lim_{x \rightarrow \infty} \frac{1}{x} = 0[/tex]

    It only makes sense to say:

    [tex]\lim_{x \rightarrow \infty} f(x) g(x) = \left( \lim_{x \rightarrow \infty} f(x) \right) \left( \lim_{x \rightarrow \infty} g(x) \right)[/tex]

    If f(x) and g(x) have limits as x approaches infinity. And as we see above -x has no limit as x approaches infinity.
  11. May 9, 2005 #10

    James R

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    Consider the following two lines:

    Line A: y = -nx
    Line B: y = x/n

    The gradients of these two lines are:

    Line A: gradient = -n
    Line B: gradient = 1/n

    Because any two lines which are perpendicular have gradients which multiply to give -1, we see immediately that line A is perpendicular to line B, for any given value of n.

    Now, consider what happens in the limit as n goes to infinity. For line B we have:

    [tex]\lim_{n \rightarrow \infty} y = \lim_{n \rightarrow \infty} x/n = 0[/tex]

    So, line B becomes the line y = 0.

    We can write line A as:

    x = -y/n

    Taking the limit as n goes to infinity, we see that this line becomes the line x = 0.

    Since we have taken the same limit in both cases, the lines A and B have remained perpendicular, and their gradients must still multiply to give -1. What we have, n terms of the gradients, is:

    [tex]\lim_{n \rightarrow \infty} (-n)(1/n) = \lim_{n \rightarrow \infty} -1 = -1[/tex]
  12. May 14, 2005 #11
    hey thats great conjecture !!

  13. May 14, 2005 #12


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    What "conjecture" are you talking about? I didn't see any conjecture in this.
  14. May 15, 2005 #13
    I Was Just Using It In The General Sense ,not Strictly In The Mathematical Sense .

    Please Ignore Whatever Does'nt Make Sense To You .
  15. May 15, 2005 #14


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    What general sense then? I thought I knew what "conjecture" meant, even in general- and I don't see how it applies. Enlighten me.
  16. May 16, 2005 #15
    ok ,you win pal !!!!
    im not gonna argue .
    tell me what it was .

    peace out ,
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