Writing equations for a line Question: Write an equation for a line containing (-8,12) that is perpendicular to the line containing the points (3,2) and (-7,2)?
What I did (this is wrong):  I found the slope for the second line, which was 0/-10. Then I changed it to be perpendicular, which was -10/0. Then I found the y-intercept by plugging in (-8,12) into the slope-intercept form and got b= -7. So, the slope intercept is y= -10/0x + 12.
This was obviously wrong, but I didn't know what to do. So, if someone can please help me, that would be great.
 A: This seems like a good point to discuss a little bit what the slope of a line means.  Consider the line $$-2x+y=6\tag{1}$$ (this is called standard form, but feel free to add $2x$ from both sides if you prefer point-slope form).

Let's try to interpret this physically.  Let's let $x$ denote the time, in seconds, from the point where we started our stopwatch and $y$ will denote the distance, in meters, a runner has traveled in that time.  So for instance, we see that the runner was already $6$ meters from the starting position when we started our stopwatch because $$-2(0) + y=6 \\ \implies y=6$$
When the stopwatch reads $2$ seconds, the runner is at $$-2(2)+y=6 \\ \implies y=10\text{ meters}$$ Likewise, the time the stopwatch reads when the runner has run $100$ meters is $$-2x + 100 = 6 \\ \implies x=47\text{ seconds}$$  We can also use this equation to backtrack a bit and figure out when the runner actually started the race -- as opposed to when we started the stopwatch.  This is when $y=0$: $$-2x+(0) = 6 \\ \implies x=-3$$ So the runner started $3$ seconds before we started the stopwatch. Hence, if this was a 100 m race, then it took the runner $47+3=50$ sec to run the whole race.
As you can see, we can get a lot of information from equation $(1)$.  But what about the speed of the runner?  Can we get that information?  Yes we can.  Recall that $\text{speed} = \dfrac{\text{distance}}{\text{time}}$.  So the speed of the runner during the 100 m race was $$\text{speed} = \frac{100 \text{ m}}{50\text{ sec}} = 2 \text{ m/s}$$
But notice that $\dfrac{\text{distance}}{\text{time}}$ is exactly the same as $\dfrac{\Delta y}{\Delta x} = \dfrac{\text{rise}}{\text{run}}$ in this case.  Hence the speed is just the slope of the line.

Let's try think to about all lines in this way -- they represent a runner running a race.  Then a greater slope means a faster runner.  A negative slope means the runner is running the wrong direction!  A slope of $0$ means the runner is protesting the race and has decided not to run at all (he's standing still).  There's one type of line where this analogy doesn't work, though.  A vertical line.  When a line is vertical how fast is the runner running?  It must be something like infinitely fast because in $0$ time he moves some nonzero distance.  This doesn't really make sense and because of this, it doesn't really make sense to talk about the slope of a vertical line -- we say the slope of a vertical line is undefined.

This is exactly the problem you're having in your exercise.  The line through $(3,2)$ and $(-7,2)$ is horizontal.  It's the line $y=2$.  So what type of line is perpendicular to a horizontal line?  A vertical line.  So you won't be able to get a slope from it -- you'll just end up dividing by $0$.  However, we can still use an equation (in standard form) to represent vertical lines.  For instance, the line through $(1,1)$ and $(1,6)$ is $x=1$.  The line through $(6,3)$ and $(6,-2)$ is $x=6$.  What should the equation of your line be?
A: As per Bye_World's comment:

First off, if you're trying to divide by 0 , you should realize you did something wrong.

No, not really. Note that the slope of the other line is $0$. The slope of this line is $\frac{1}{0}$. This means that this line is vertical, as it should be, given that it's perpendicular to a horizontal line.
All vertical lines can be written as $x=c$, where $c$ is a constant. All you have to do here is figure out what $c$ is.
