Explain why perpendicular lines have negative reciprocal slopes I am not sure how to explain this. I just know they have negative reciprocals because one one line will have a positive slope while the other negative. 

Assuming experience with algebra without calculus background. So, I would suggest keeping to the idea that slope, $m$, is equal to "rise over run." Given a line with slope, lets say $\frac{a}{b}$, that means it rises $a$ in the $y$ direction for every $b$ it goes in the positive $x$ direction in the plane. (I would also encourage to always keep $b>0$ so we always go in the positive $x$ direction.) The way to make the slope the "most opposite" is to flip it and make it negative.  Now, that is nowhere close to a proof, but I found one that only uses the fact that the Pythagorean theorem is true when you have a right angle and high school algebra. 
Claim: If two lines in the plane, $f(x)=mx+b$ and $g(x)=nx+c$, are perpendicular, then $n=\frac{-1}{m}$.
Two lines $f(x)=mx+b$ and $g(x)=nx+c$ are not parallel, so they intersect.  Assume that they do not intersect on the $y$-axis, i.e. $c \neq   b$. Then the triangle formed by the graphs of these two lines and the $y$-axis is a right triangle if the Pythagorean theorem holds. WLOG, assume that $c>b$. The lengths of the side on the $y$-axis will be $c-b$. To find the other two sides, we do a little algebra. The intersection of these lines is $mx+b=nx+c$ and solving for $x$, we find that $x=\frac{c-b}{m-n}$. Thus, the point of intersection is
$$\left(\frac{c-b}{m-n},\frac{m(c-b)}{m-n}+b\right)=\left(\frac{c-b}{m-n},\frac{n(c-b)}{m-n}+c\right)$$
which we get by plugging in our $x$ for $f$ and $g$. To find the distance of each of the two legs of our triangle, we just use the distance formula, and find the the distance from
$$\left(\frac{c-b}{m-n},\frac{m(c-b)}{m-n}+b\right) \text{ to } (0,b) $$ 
is $\frac{(c-b)\sqrt{1+m^2}}{m-n}$.  For the other side, we use the distance from 
$$\left(\frac{c-b}{m-n},\frac{n(c-b)}{m-n}+c\right) \text{ to } (0,c) $$ 
which is $\frac{(c-b)\sqrt{1+n^2}}{m-n}$. Now, we set up the Pythagorean theorem, which we can use since the angle between the lines is right, and see that 
$$ (c-b)^2 = \left[\frac{(c-b)\sqrt{1+m^2}}{m-n}\right]^2 +  \left[\frac{(c-b)\sqrt{1+n^2}}{m-n}\right]^2 $$
Canceling the $(c-b)^2$ and multiplying by $(m-n)^2$ to both sides, we get 
$$(m-n)^2 = 1+m^2 + 1+n^2$$ 
$$m^2 -2mn + n^2 = 2+m^2 +n^2 $$ 
Canceling and simplifying, we find that $n=\frac{-1}{m}$. 
