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.
Translate two lines so that their intersection is the origin and then take two vectors along each line, say $u=(1,k_1), v=(1,k_2)$. The two lines are perpendicular if and only if $u\perp v$, viz
$$u\cdot v=1+k_1k_2=0$$
This explains why $k_1$ is the negative reciprocal of $k_2$.
Two 90 degree rotations will snap back to how it was originally:
Suppose you have line A, line A' which is perpendicular to A, and line A'' which is perpendicular to A' (all on the Cartesian plane). A and A'' are parallel (since A'' is A rotated ±90±90 degrees, and even the ± is a silly distinction as rotating 90 degrees clockwise is the same as rotating 90 degrees counterclockwise), which means that if perpendicular slopes have a hard and fast formula, they had better be reciprocals or negative reciprocals (or the same slope, but that's garbage), as otherwise p(p(m)), the perpendicular of the perpendicular of the slope, will not equal the original slope. By inspection, in a pair of perpendicular lines, one has a positive slope and one has a negative slope (I'm not account for the student who asks "What about zero?"). Then the only valid option is negative reciprocals.

In my opinion, this is the best way to show a middle school student that will allows them to teach it to other middle school students (but geometry/trigonometry should eventually show it formally). Rotating a piece of paper 90 degrees and describing how it transforms rise and run is my second favorite.