Why is the product of 2 perpendicular gradients = -1?

Christophicus

Is there a proper proof, or is just defined as this?

Reply 1

ArVi

Widowmaker

Is there a proper proof, or is just defined as this?

The formula for the angle θ between two lines with gradients m1 and m2:

tan θ = | (m1-m2) / (1 + m1m2) |

so when θ = 90, (perpendicular)
tan θ is undefined

since LHS is undefined, RHS should also be undefined.

for RHS to be undefined, 1 + m1m2 = 0 (so it gets divided by 0)

rearranging gives: m1m2 = -1

Reply 2

-G-a-v-

Alternatively...

Pick a point P(a, b) in the first quadrant of a graph, and join it up to the origin O. Then in the second quadrant, draw a line from the origin passes through the point Q(-b, a). Let the gradient of OP be m1, and the gradient of OQ be m2

so..

m1 = b/a
m2 = -a/b

so m1 x m2 = b/a x -a/b
= -ab/ab
= -1

It'll be clearer if you draw a diagram but thats the basic idea.

Reply 3

john!!

I like the second proof more... "undefined" isn't very rigorous, for example, "undefined" = "undefined" => "1/0 = 5/0" => "1 = 5"

Reply 4

AlphaNumeric

john!!

"undefined" = "undefined" => "1/0 = 5/0" => "1 = 5"

That is an operation you cannot do. 1/0 is not a number, in the same way "blue" is not a number.

Most of the kind of "dodgy" maths you see which prove 1=2 rely in a slight of hand to hide the fact they are dividing by zero. Since 0 has no multiplicative inverse (ie 1/0 doesn't exist) you cannot do that operation, or you could just apply it to 1*0 = 2*0 and you'd get 2=1.

There are rigorous ways to deal with the limit of 1/x as x->0, and other such functions, there are infact entire branches of mathematics, but they involve things like complex numbers.

Reply 5

john!!

Yeah that was the point I was trying to make. That was an example of it not being rigorous.

Reply 6

Cexy

It was still (kind of) rigourous, given that for every value of m₁m₂ except m₁m₂ = -1, the right hand side was finite. Since you knew that the left hand side was undefined, your only choice is to take m₁m₂ = -1.

But I agree that distortedgav's proof was better.

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