The question is more for mathematic thrillers. as u know if two lines are perpendicular the product of there?

Question

slope is (-1). Now if we take X-axis & Y-axis slope of X-axis is 0 and that of Y-axis is infinity or is say indeterminant then product of these slope is also indeterminant. then how can u say that they are perpendicular. Don't tell me to measure the angle. i want to know why is that contradiction is here. i hope i was able to express my self.

I think i was not able to express myself.
now in mathematics a theorem is mostly prove by induction.and only one example is sufficient to throw that theorem out. now the product of two perpendicular lines is -1 is a proved theorem or i say rule but the contradiction is on the X&Y- axis there product of slope doesn't turn out to be -1 now why is that so there must be a reason given for it or an assumption made for it i want to know that thing.
and lastly the "?" mark was putted by default by yahoo's read it with a pace without "?" mark.

Answer 1

The problem stands with your initial assertion. The correctly stated property isn't:
- The product of the slopes of perpendicular lines is -1.
It's:
- The slopes of perpendicular lines are the negative inverses of each other.

It just so happens that multiplying negative inverses gives you -1, EXCEPT when one of the numbers is 0.

Answer 2

Consider the following limiting process, of the rotation of two definitely perpendicular lines Ox' and Oy', which are initially both VERY close to the corresponding axes Ox and Oy. ('O' stands for the origin.)

Let tan (x'Ox) = - epsilon, epsilon a small positive number. Then Ox' points slightly downwards from the direction Ox, essentially at an angle of - epsilon radians. Since the product of the slopes of perpendicular lines is -1, Oy' has slope tan (y'Ox) = 1/epsilon. This being extremely large, it is an initially almost vertical direction, still in the first quadrant of the (x, y) coordinate system.

Now rotate the PAIR of primed axes anti-clockwise, keeping them always perpendicular to one another, so that epsilon ---> 0.

As this process takes place, tan (x'Ox) ---> tan (xOx) = 0, and tan (y'Ox) ---> tan (yOx). But at each step of the latter process, tan (y'Ox) is still precisely 1/epsilon, with the product of slopes for the primed pair of axes -1. So the limiting process is perfectly well defined, with each of their values linked by this finite and constant product. Just as 0 is a perfectly well-defined limit for the process epsilon ---> 0, so is that linked "+ infinity" a well-defined limit for the process lim (1/epsilon) as epsilon ---> 0.

The difficulty that you claim to present lies in your loose assertion that the slope of the y axis IS infinity (or indeterminate) and therefore (triumphantly) that "the product of these slopes is also indeterminate.":

NOT SO! You have inverted the direction of the valid logical argument, and that is impermissible. The product of the slopes of perpendicular lines is ABSOLUTELY FIXED, at - 1; but if the value of one of them tends to zero, the other necessarily tends to infinity. One can NOT reverse this argument to assert that the product must be indeterminate. It was held determinate throughout the limiting process, and no amount of verbal legerdemain can justifiably claim, and certainly not prove, otherwise.

BAH, HUMBUG! Nevertheless:

Live long and prosper.

Answer 3

How do you know the rule is true?
Any proof of the rule excludes lines of indeterminate slope. Therefore, this rule simply does not apply to the x- and y-axes.

Answer 4

hmm this is a good question.. -1/0 approaches -infinity; it seems as reasonable for the y axis to have a slope of - infinity as of infinity.