# Product Of Slopes Of X,y Axes=0

1. May 6, 2005

### extreme_machinations

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

2. May 6, 2005

### Zurtex

It's funny how both the y-axis and the line y=x have the same gradient according to you :tongue:

3. May 8, 2005

### extreme_machinations

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

4. May 8, 2005

### Zurtex

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.

5. May 8, 2005

### honestrosewater

This reminds me of the "proof" that 1 = 2. The mistake involved dividing by zero or something. ;)

6. May 8, 2005

### HallsofIvy

Staff Emeritus
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!

7. May 8, 2005

### James R

The slope of the y axis (x=0) is infinite. And, as we all know, infinity times zero equals negative 1. :)

8. May 9, 2005

### extreme_machinations

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 ??

9. May 9, 2005

### Zurtex

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:

$$\lim_{x \rightarrow \infty} -x \frac{1}{x} = -1$$

Now although it's true that:

$$\lim_{x \rightarrow \infty} -x = -\infty$$

And:

$$\lim_{x \rightarrow \infty} \frac{1}{x} = 0$$

It only makes sense to say:

$$\lim_{x \rightarrow \infty} f(x) g(x) = \left( \lim_{x \rightarrow \infty} f(x) \right) \left( \lim_{x \rightarrow \infty} g(x) \right)$$

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.

10. May 9, 2005

### James R

Consider the following two lines:

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

The gradients of these two lines are:

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:

$$\lim_{n \rightarrow \infty} y = \lim_{n \rightarrow \infty} x/n = 0$$

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:

$$\lim_{n \rightarrow \infty} (-n)(1/n) = \lim_{n \rightarrow \infty} -1 = -1$$

11. May 14, 2005

### extreme_machinations

hey thats great conjecture !!

thanks

12. May 14, 2005

### HallsofIvy

Staff Emeritus

What "conjecture" are you talking about? I didn't see any conjecture in this.

13. May 15, 2005

### extreme_machinations

I Was Just Using It In The General Sense ,not Strictly In The Mathematical Sense .

Please Ignore Whatever Does'nt Make Sense To You .

14. May 15, 2005

### HallsofIvy

Staff Emeritus
What general sense then? I thought I knew what "conjecture" meant, even in general- and I don't see how it applies. Enlighten me.

15. May 16, 2005

### extreme_machinations

ok ,you win pal !!!!
im not gonna argue .
tell me what it was .

peace out ,