# Is the slope of the y axis infinity?

Dec 14, 2016

When we derive this result, we take the slope of one line as
$\tan Q$ and the slope of other line (Perpendicular to it) as

$\tan \left(90 + Q\right) = - \cot Q$

hence their product becomes -1 .

For your case the slopes are$\tan 90$ and $- \cot 90$ the product in this case is -1 .

Also the product of a number tending to infinity and a number tending to zero is not fixed and it depends upon the question .

Dec 14, 2016

No. The slope of the $y$ axis is best considered "undefined".

#### Explanation:

The slope of the $x$ axis is $0$.

The slope of the $y$ axis is undefined.

You can try hard to make it "infinity", but what "infinity" do you mean?

For example an standard calculus definition would give you:

${\lim}_{x \to 0 +} \frac{1}{x} = + \infty$

${\lim}_{x \to 0 -} \frac{1}{x} = - \infty$

So using these kind of definitions, you would not know if the slope of the $y$ axis is $+ \infty$ or $- \infty$.

Note that $+ \infty$ and $- \infty$ do not behave like proper numbers. You cannot perform many arithmetic operations on them.

For example:

What is $\infty - \infty$ ?

What is $0 \cdot \infty$ ?

Both are indeterminate.

The property that the product of the slopes of a pair of perpendicular lines is $- 1$ holds when the slope of both lines is determined, but the slope of the $y$ axis is not.

Intuitively, the slope of the $y$ axis is some kind of infinity. Is there something we can do?

Instead of the standard calculus objects $+ \infty$ and $- \infty$, you can add just one "infinity" to the Real line to get something called the projective Real line ${\mathbb{R}}_{\infty}$. Then you can define $\frac{1}{0} = \infty$ and $\frac{1}{\infty} = 0$, but $0 \cdot \infty$ is still indeterminate.