Neat slope trick
We discovered (I assume rediscovered, since all of this stuff has been done before, and better) a neat way to show something about slopes.
Plot a right triangle with vertices at (0,0), (0,b) , and (-a,b). Notice that the slope of the hypotenuse is -b/a.
Plot another right triangle with vertices at (0,0), (0,a), and (b,a). Notice that the slope of the hypotenuse is a/b.
Also notice that since the triangles are congruent, the sum of the angles that meet at the origin is 90, and that the lines are perpendicular.
We are only a bunch of really nitpicky details away from a nice proof that lines in the plane that cross both axes are perpendicular if and only if the product of their slopes is -1.
Does anyone know if this is a widely used demonstration?
JD2718 
Nice! I like coordinate geometry proofs. I guess all you’d have to do to pick those nits is generalize away the origin. And maybe set up something to explain how you build the right triangles around two lines, as opposed to building the lines from these convenient triangles. But I have always been at a loss to explain simply and quickly why the product of slopes of perpendicular lines is -1. Here already you have a fast demonstration. Thanks!
Wow! This is a really nice way to demonstrate that perpendicular lines are negative reciprocals of each other. It may not be widely used, but it will be “widely used” by me.
I have never taught my students that the product of the slopes of perpendicular lines equals -1, I’ve just relied upon the students recognition of negative reciprocal. I think your way is much easier for students to recognize and understand.
I will be incorporating both into my lesson this year.
If either of you have any luck with this, good or bad, would you mind sharing? I have an off-track geometry, and we will begin coordinate proofs in two weeks. When I have some live classroom feedback, I will share it here as well.