Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

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

    • Thread Starter
    Offline

    12
    ReputationRep:
    If you had a line with gradient m, how would you prove that the gradient of the normal to this line is -\dfrac{1}{m}?

    Thanks in advance


    Posted from TSR Mobile
    Offline

    16
    ReputationRep:
    (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 -\dfrac{1}{m}?

    Thanks in advance


    Posted from TSR Mobile
    I would write the lines in vector form and use the dot product = 0
    Offline

    2
    ReputationRep:
    You could also do \displaystyle \left(\begin{matrix}0&-1\\1&0\end{matrix}\right) \left(\begin{matrix}m\\1\end{mat  rix}\right) = \left(\begin{matrix}-1\\m\end{matrix}\right).

    Since the vector \left(\begin{matrix}m\\1\end{mat  rix}\right) corresponds to that of a line with gradient m, rotating it by 90 degrees creates a vector \left(\begin{matrix}-1\\m\end{matrix}\right), which corresponds to a line with gradient \displaystyle -\dfrac{1}{m}.
    Offline

    18
    ReputationRep:
    (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 -\dfrac{1}{m}?

    Thanks in advance


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


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (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.
    Posted from TSR Mobile
    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
    Offline

    2
    ReputationRep:
    I once saw a geometric proof but can't for the life of me recall it at this moment in time
    Offline

    11
    ReputationRep:
    (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.
    Offline

    10
    ReputationRep:
    (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.
    Offline

    18
    ReputationRep:
    (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.


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by physicsmaths)
    Lol na. Much much easier.
    Posted from TSR Mobile
    ok
    plz tell. you promised :cool:
    Offline

    18
    ReputationRep:
    (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


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by physicsmaths)
    ill do it now. its to do with symmetry. i have a feeling it might be wrong though lol
    Posted from TSR Mobile
    cool
    Offline

    18
    ReputationRep:
    (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.


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by physicsmaths)
    Notice symmetry with two x values and tw y values... it should be obvious from there
    Posted from TSR Mobile
    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
    Offline

    18
    ReputationRep:
    (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.


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by physicsmaths)
    lol. Label the points. U shud see the gradients involve all the same points.
    Posted from TSR Mobile
    i lol u back m8
    honestly bro ur talking gibberish- plz enlighten us some more
    maybe a pic perhaps?
    Offline

    18
    ReputationRep:
    (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.


    Posted from TSR Mobile
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 19, 2015
Poll
Do I go to The Streets tomorrow night?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.