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

1. If you had a line with gradient m, how would you prove that the gradient of the normal to this line is ?

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2. (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 ?

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I would write the lines in vector form and use the dot product = 0
3. You could also do .

Since the vector corresponds to that of a line with gradient , rotating it by 90 degrees creates a vector , which corresponds to a line with gradient .
4. (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 ?

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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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5. (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
6. I once saw a geometric proof but can't for the life of me recall it at this moment in time
7. (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.
8. (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.
9. (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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10. (Original post by physicsmaths)
Lol na. Much much easier.
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ok
plz tell. you promised
11. (Original post by MathMeister)
ok
plz tell. you promised
ill do it now. its to do with symmetry. i have a feeling it might be wrong though lol

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12. (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
13. (Original post by MathMeister)
cool
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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14. (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

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
15. (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

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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16. (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
maybe a pic perhaps?
17. (Original post by MathMeister)
i lol u back m8
honestly bro ur talking gibberish- plz enlighten us some more
maybe a pic perhaps?
lol my phone is being a bi*ch. i aint got no memory bro.

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