*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

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!

C

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.