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How would you prove that the gradient of the normal is -1/m?

If you had a line with gradient m, how would you prove that the gradient of the normal to this line is −1m-\dfrac{1}{m}?

Thanks in advance


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(edited 9 years ago)
Original post by Brian Moser
If you had a line with gradient m, how would you prove that the gradient of the normal to this line is −1m-\dfrac{1}{m}?

Thanks in advance


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I would write the lines in vector form and use the dot product = 0
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You could also do (0−110)(m1)=(−1m)\displaystyle \left(\begin{matrix}0&-1\\1&0\end{matrix}\right) \left(\begin{matrix}m\\1\end{matrix}\right) = \left(\begin{matrix}-1\\m\end{matrix}\right).

Since the vector (m1)\left(\begin{matrix}m\\1\end{matrix}\right) corresponds to that of a line with gradient mm, rotating it by 90 degrees creates a vector (−1m)\left(\begin{matrix}-1\\m\end{matrix}\right), which corresponds to a line with gradient −1m\displaystyle -\dfrac{1}{m}.
Original post by Brian Moser
If you had a line with gradient m, how would you prove that the gradient of the normal to this line is −1m-\dfrac{1}{m}?

Thanks in advance


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i have an extremely easy way. very very easy. i shall post it tmmrw though as im very tired right now.


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Original post by physicsmaths
i have an extremely easy way. very very easy. i shall post it tmmrw though as im very tired right now.
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I always wondered this question and tried to figure it out myself.
Is it enough to say tan theta = gradient since tan=sin/cos
Then tan x+ pi upon 2 radians = -cotx = -1/tanx=-1/m?
You can also prove this with the addition formulae---> which are also easily proven lol
I once saw a geometric proof but can't for the life of me recall it at this moment in time
Original post by GorlimtheUnhappy
I once saw a geometric proof but can't for the life of me recall it at this moment in time


It can be demonstrated geometrically by constructing a pair of similar triangles at the point of intersection, together with the idea that for a straight line, "increasing x, decreasing y", implies a negative gradient.

For the similar triangles, the "rise" of one turns into the "run" of the other, and vice versa.
(edited 9 years ago)
Original post by atsruser
It can be demonstrated geometrically by constructing a pair of similar triangles at the point of intersection, together with the idea that for a straight line, "increasing x, decreasing y", implies a negative gradient.

For the similar triangles, the "rise" of one turns into the "run" of the other, and vice versa.


You can use tracing paper for a quick visual demonstration of this.
Original post by MathMeister
I always wondered this question and tried to figure it out myself.
Is it enough to say tan theta = gradient since tan=sin/cos
Then tan x+ pi upon 2 radians = -cotx = -1/tanx=-1/m?
You can also prove this with the addition formulae---> which are also easily proven lol


Lol na. Much much easier.


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Original post by physicsmaths
Lol na. Much much easier.
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ok
plz tell. you promised :cool:
Original post by MathMeister
ok
plz tell. you promised :cool:


ill do it now. its to do with symmetry. i have a feeling it might be wrong though lol


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Original post by physicsmaths
ill do it now. its to do with symmetry. i have a feeling it might be wrong though lol
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cool :biggrin:
Original post by MathMeister
cool :biggrin:


Take two perpendicular lines. Construct a square around the intersection. Notice symmetry with two x values and tw y values. It's all pretty straight forward from there. I'm not fully sure it would suffice. Ur it should because it is a straight line and rate of change for big x big y would be the same for and x n any y.


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Original post by physicsmaths
Notice symmetry with two x values and tw y values... it should be obvious from there
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lolwut
I drew 2 lines intersecting and a square around the point of intersection.
you didn't quite make sense bro with the last bit :redface:

in addition the formula is rather intuitive anyway...
considering angles from 0-90 take m, or theta to be large then -1/m would give a negative gradient which is very small (which makes sense since given the formula and by rotating the line by 90 degrees (the line was already at a small angle from the vertical))
its not hard to think about rly
(edited 9 years ago)
Original post by MathMeister
lolwut
I drew 2 lines intersecting and a square around the point of intersection.
you didn't quite make sense bro with the last bit :redface:

in addition the formula is rather intuitive anyway...
considering angles from 0-90 take m, or theta to be large then -1/m would give a negative gradient which is very small (which makes sense since given the formula and by rotating the line by 90 degrees (the line was already at a small angle from the vertical))
its not hard to think about rly


lol. Label the points. U shud see the gradients involve all the same points.


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Original post by physicsmaths
lol. Label the points. U shud see the gradients involve all the same points.
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i lol u back m8
honestly bro ur talking gibberish- plz enlighten us some more :tongue:
maybe a pic perhaps?
Original post by MathMeister
i lol u back m8
honestly bro ur talking gibberish- plz enlighten us some more :tongue:
maybe a pic perhaps?


lol my phone is being a bi*ch. i aint got no memory bro.


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