Why $\infty×0=-1$ from multiplication of two slopes of two lines perpendicular to each other and how do we define infinity? 
Here is given $A(x_1,y_1), B(x_1,y_2), C(x_2,y_3)$ and $D(x_3,y_3)$. I have recently read that, multiplication of two perpendicular lines is always $-1$.
From the above graph, the slope of $AB, m_1 = \frac{y_2-y_1}{x_1-x_1} = \frac{y_2-y_1}{0} = \infty$
and the slope of $CD, m_2 = \frac{y_3-y_3}{x_3-x_2} = \frac{0}{x_3-x_2} = 0$
And then we get $m_1m_2 = -1$
$\infty*0 = -1$
Here it is clearified that multiplication of infinity and zero always leads to $-1$ from the above formula. Then in which fact, do we strongly put emphasis on? $\infty*0 = -1$ or $\infty = \frac{-1}{0}$
If it indicates the 1st condition, then why $\infty*0$ would be only equal to $-1$. It could have been resulted in any real number. And how we define the infinity in this case?
Any kind of conception would be greatly helpful for me to remove my ignorance. Thanks in advance.
 A: Good question! The trick is: infinity is not a number (in most frameworks).
When we write ∞, we're using it as a convenient shorthand for "what happens when this variable gets arbitrarily large". And if you try to use it as a number, things go horribly wrong.
When you ask about something like $0 \times \infty$, what you're usually asking is: "what happens to $0 \times k$, as $k$ gets arbitrarily large?" Using limit notation, we'd say "what is $\lim_{k \rightarrow \infty} 0 \times k$?"
And sometimes this can be answered: we can see that, no matter how big $k$ gets, $0 \times k$ is always zero. So $\lim_{k \rightarrow \infty} 0 \times k = 0$. Sometimes it can't. Suppose we instead asked about $\frac{1}{k} \times k^2$: the first part is going to zero, the second part is going to infinity. But this time, as $k$ gets bigger and bigger, the product just keeps getting bigger and bigger too. So in this case, our "$0 \times \infty$" ends up being infinite.
So the answer to $0 \times \infty$ is: "it depends how you get there". Infinity isn't a number, and it doesn't act like one: rather, it's a shorthand for "let's see what happens when we let this variable get arbitrarily large".
A: The answer by Haran is incorrect, and the answer by Draconis is a bit unclear. Specifically, one cannot say that "$0·∞$ can take any value". In the first place, "$1/0$" is ill-defined when $1,0$ are ordinary reals, so there is no way to get any such thing as a real $\infty$. Thus "$0·∞$" is simply a meaningless expression unless you give a rigorous definition of "$∞$" as some mathematical object. For example, you could very well let $∞ = 8$ even if it is unconventional.
Of course, in real analysis we often have limits to infinity, but that is not the same as just doing an illegal division by zero. We normally define "$\lim_{x→c} \cdots = ∞$" as merely notational short-hand for some formal statement, and not that the limit expression has a value that is equal to something called $∞$. That is why we cannot write "$\lim_{x→0} \cos(1/x) ≠ ∞$" because in the first place "$\lim_{x→0} \cos(1/x)$" is undefined.
Also, although it is true that $\lim_{x→0} 1/x^2 = ∞$, it is meaningless to write "$1/0^2 = ∞$" (unless you're not working with ordinary reals, in which case you must know precisely what you're doing).
Draconis' answer is unclear, because you must very clearly understand that "$0·∞$" is not an expression with a value. Rather, it is merely a string of symbols. (See this related post about the difference between $0^0$ and the form "$0^0$".)

As for your initial question, your mistake is right at the beginning; your equation for a vertical line is wrong. In general, the equation of a line in the $x,y$-plane is { $ax+by=c$ } for some reals $a,b,c$. You cannot divide the equation by $b$ if $b=0$. So you cannot even get to the point of 'computing' the slope of a vertical line.
But there are correct versions of the facts you're interested in:

Given any line $L$ in the $x,y$-plane with equation { $ax+by=c$ }, the slope of $L$ is the ratio $-a:b$.
Given any perpendicular lines $L,M$, their slopes $a:b$ and $c:d$ satisfy $a:b = -d:c$.

Note that $1:0$ and $0:1$ are both valid ratios! "$0:0$" is not a valid ratio. (At least one must be nonzero.) Two ratios $a:b$ and $c:d$ are equal iff $ad = bc$. In particular, notice that $1:0 = -1:0$.
Also note that this formulation handles all kinds of lines correctly (without facing illegal division by zero).
A: When you take the limit for your answer at these expressions, it happens that the limit tends to $-1$. There is nothing special about the value $-1$ itself. Consider $n \cdot \frac{2}{n}$ where $n$ tends to infinity. The value is always $2$. However, this does not mean you can directly set $n= \infty$ for a proper answer. Expressions like $0 \cdot \infty$ are not undefined expressions which have no answer, but instead, indeterminate forms, which can take any answer (limit) based on the situation.
