Question

Is the slope of the `y` axis infinity?

Answer 1

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(90+Q) = - cotQ`

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 .

Answer 2

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->0+) 1/x = +oo`

`lim_(x->0-) 1/x = -oo`

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

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

For example:

What is `oo - oo` ?

What is `0 * oo` ?

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 `+oo` and `-oo`, you can add just one "infinity" to the Real line to get something called the projective Real line `RR_oo`. Then you can define `1/0 = oo` and `1/oo = 0`, but `0 * oo` is still indeterminate.