Vertical Lines & Horizontal Lines and Their Slopes Hello, in my discrete mathematics for Information Technology, my instructor asked this question of us:
Discuss in your own words if it makes sense that a vertical line has an infinite slope and a horizontal line has a slope of zero.

My answer is:

The slope of a vertical line is undefined.

The slope of a horizontal line is zero.

Think about what slope is and it will make sense. Slope is rise over run and and be found by using 2 points on the line, (x1,y1) and (x2,y2)

slope = (y2 - y1) / (x2 - x1)

On a horizontal line all the y values are the same so y2-y1 will always be zero so the slope will be zero regardless of what x2-x1 equals.

On a vertical line all the x values are the same so x2-x1 will always be zero. Dividing by zero is undefined so the slope is undefined. 

My brother has told me this is incorrect, but refuses to explain it. Can someone please confirm this?
It's a good question from your instructor because it prompts you to think about infinity, and by thinking about infinity sufficiently precisely you can argue for both answers.

If this is to make any sense, it needs to be considered as a limiting case. But a vertical line is a limiting case of both a line with slope $+\infty$ and a line with slope $-\infty$. For instance, we can make this precise by considering a line with slope $m$ given by

$$\vec r(s)=\vec r_0+\frac s{\sqrt{1+m^2}}\pmatrix{1\\m}\;,$$

where $s$ is the distance along the line. If we take $m$ to $\pm\infty$, $\vec r(s)$ converges pointwise to $\vec r_0\pm s\vec e_z$, so the locus of the limits is a vertical line. Thus a vertical line can equally well be described by a slope of $+\infty$ or a slope of $-\infty$, so it makes no sense to say that it has slope $+\infty$.

But who said that "infinity" means $+\infty$? Consider the slope represented not on the real line, by the number $m$, but on the unit circle, by the point at which the line through $(0,m)$ and $(1,0)$ meets the unit circle. Now taking $m$ to $+\infty$ or $-\infty$ leads to the same limit point $(1,0)$. Nothing forces us to add two different infinities to the real line, one at each "end". Though this is useful for many purposes, it's not useful when we consider the real line as representing the set of possible slopes. The construction indicates how we can consistently add a single infinity to that set and consider both $m\to+\infty$ and $m\to-\infty$ as the same limit where the slope, represented on the unit circle, goes to the point $(1,0)$. In this sense, it makes sense to say that a vertical line has infinite slope.
If a line had an undefined slope I would not be able to draw it.  I can draw a vertical line.  It must therefore have a defined slope.