The slope of $x$-axis is $0$ while that of $y$-axis is $\tan 90$,infinity. In coordinate geometry, the product of slope two perpendicular lines is $-1$. Does that apply to the coordinate axes as well, so is '$\infty\cdot 0 = -1$'?

  • $\begingroup$ To begin with, the $y$-axis doesn't have a slope. Some people say that it's infinity, but that is not true, although it makes some sense if taken informally. $\endgroup$ – user384138 Dec 16 '16 at 20:27

You are right in the terms of first thing you said which is:

Product of Slope of two perpendicular lines is $-1$. Here slope is just the tangent of the anle given straight line make with positive $x-$axis.

Now, For the $x-$axis, the slope is $0$ as the angle between $x-$axis and $x-$axis is $0$ and $\tan 0=0$. For the same reason the slope of $y-$axis is $\tan 90=$ Not defined. People can have good arguments and discussion can go forever, but I shall avoid the term infinity here and use the term not defined.

Now whenever this term not defined$(\frac{0}{0})$ comes in mathematical context, it has only one purpose which is to inhibit us to proceed. When you use $(\frac{0}{0})$, you lose every right (From right I mean right to do arithmetic operation).

So, trying to compute $\frac{0}{0}\times 0$ is ultimately illegitimate in mathematics. Hence answer of your second Question is $NO$.


No. The statement "The product of the slopes of two perpendicular lines is $-1$" applies only to lines whose slope is neither $0$ nor $\infty$; "$0\cdot \infty$" is undefined. (Such weird entities arise, sort of, in calculus as indeterminate forms, but even then there isn't a single value of "$0\cdot\infty$".)

  • $\begingroup$ So you mean to say that the condition of perpendicularity does not apply to the coordinate axes. Do we ignore them because of such an 'o*infinity' form or is the condition of being perpendicular based on an assumption that is not satisfied by the axes. Also does 'there is not a single value' imply the existence of many values ? $\endgroup$ – Soumyarup Choudhury Dec 16 '16 at 20:35
  • $\begingroup$ The quickest answer to your question is that you cannot treat $\infty$ as a standard real (or complex) number. $\endgroup$ – Michael McGovern Dec 16 '16 at 21:02
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    $\begingroup$ The condition of perpendicularity does apply to the coordinate axes. The point is that perpendicularity is not equivalent to "the product of the slopes equals $-1$". Perpendicularity is a geometric property. Sometimes we can detect perpendicularity by a nice arithmetic formula, sometimes we cannot. The full statement is that two lines $L,M$ are perpendicular if and only if one of two things happens: (1) one of $L,M$ is horizontal and the other is vertical; or (2) neither is vertical and the product of their slopes equals $-1$.@SoumyarupChoudhury $\endgroup$ – Lee Mosher Jan 4 '17 at 2:25
  • $\begingroup$ @LeeMosher My apologies, I misunderstood (and I've deleted my previous comment). $\endgroup$ – Noah Schweber Jan 4 '17 at 2:32

From just tan(0 degrees) and tan(90 degrees), horizontal and vertical, you can't just multiply 0 x $\infty$ and get 1.

Keep in mind $\tan(x) = \frac{\sin(x)}{\cos(x)}$.

The product of a horizontal and vertical line would be $\tan(0)\tan(\frac{π}{2})$, where the angles are expressed in radians.

This is equal to $\frac{\sin(0)}{\cos(0)} * \frac{\sin\left(\frac{π}{2}\right)}{\cos\left(\frac{π}{2}\right)}$.

This is equal to $\frac{0}{1} * \frac{1}{0}$ which could arguably be equal to 1, but this is flawed.

To prove something like this, you would use a limit, in where you find $\lim_{x\to0}\tan(x)\tan\left(x+\frac{π}{2}\right)$.

However still, because it is an indeterminate form, the property of perpendicular lines' slopes having a product of 1 only applies to lines with real slopes.


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