Dear Uncle Colin,

I don’t understand why the normal gradient is the negative reciprocal of the tangent gradient. What’s the logic there?

-- Pythagoras Is Blinding You To What’s Obvious

Hi, PIBYTWO, and thanks for your message!

My favourite way to think about perpendicular gradients is to imagine a line going across a chessboard. Let’s say it’s $y = 3x + 5$, just to start with a simple one.

Every square the line goes to the right, it goes up three. That’s what having a gradient of 3 means. You can imagine a triangle on the underside of the line, one unit wide, three units tall. Got that picture in your head? Good.

Now rotate that triangle through a right angle. Doesn’t really matter which way, or about which point, but I’d take the bottom-left corner and rotate it clockwise, so that it’s below the line heading down and to the right: the “1” side is now on the left and the “3” side is now on the bottom. The line is going one unit down for every three it goes across, so its gradient is $\frac{-1}{3}$. Lookit! A negative reciprocal!

Now, there’s nothing special about the “3” in that triangle - it could just as well have been “2” or “12” or “$\frac{3}{2}$” or “$\pi$” or even - let’s get wildly abstract - “$m$”.

In that last case, the rotated triangle defines a line with gradient of $\frac{-1}{m}$ - which is the negative reciprocal of the original gradient.

Hope that helps!