Is the slope of the #y# axis infinity?

Question

Allison Allen · Accepted Answer

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 .

James Atkins · Answer

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

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.

Sadie Adams · Answer

undefined No, the slope of the y-axis is undefined, not infinity.