1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Product of the gradients of perpendicular lines proof help

  1. Apr 19, 2015 #1
    Okay I'm having a little trouble understanding a section of this proof about the product of the gradients of perpendicular lines given in my textbook. I'm gonna type the proof out but there will be a link at the bottom to an online version of the textbook so you can see the accompanying diagram too.

    I understand essentially the whole proof except this one thing it says at the beginning. This bit: 'If AB makes an angle theta with the x-axis then CD makes an angle theta with the y axis.'

    So my question is this: Why do the x and y axis make the same angle (theta) when they intersect AB and CD? What's the reason/proof behind that?

    Will be very grateful if anyone can explain this clearly to me:)

    Here is the link to my textbook (go to slide 79 - the bit on perpendicular lines at the bottom): http://www.slideshare.net/mobile/Slyscott12/core-maths-for-a-level-3rd-edition-by-lbostock-schandler
    Here is the proof typed out:
    Consider the perpendicular lines AB and CD whose gradients are m1 and m2 respectively.
    If AB makes an angle theta with the x axis then CD makes an angle theta with the y axis. Therefore triangles PQR and PST are similar.
    Now the gradient of AB is ST/PS=m1
    And the gradient of CD is - PQ/QR=m2,i.e PQ /QR= - m2

    But ST/PS=QR/PQ (since triangles PQR and PST are similar)

    Therefore m1= - 1/m2 or m1 *m2 = - 1
  2. jcsd
  3. Apr 19, 2015 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    BPD is a right angle. (Given)

    SPQ is a right angle because PS is parallel to the x-axis and QP is parallel to the y-axis.

    Now consider angle SPR...
  4. Apr 19, 2015 #3


    User Avatar
    Gold Member

    1- AB makes an angle θ with y axis. If both lines are rotated 90 deg cw, then AB is now parallel with CD and y axis is now parallel with x axis, so angle between CD and x axis is also θ.

    2 - Consider the quadrilateral APDO. The corners at O and P are 90 deg by definition. The 4 corners must total 360 , so corners at A and D must total 180. At corner A, the acute angle and obtuse angle with y axis must also total 180. So the acute angles between the lines and the axes must be equal, and the obtuse angles between the lines and the axes must be equal.
  5. Apr 19, 2015 #4
    Okay I think I might have it. So going from what you said:

    ∠ BPD=90° and ∠SPQ=90°

    From the diagram ∠BPS+∠SPR =∠BPD
    then ∠BPS +∠SPR = 90°

    So ∠BPS = 90°-∠SPR

    Also, as ∠SPR +∠RPQ =∠SPQ then

    ∠SPR +∠RPQ = 90°

    So ∠RPQ = 90° - ∠SPR

    Therefore ∠RPQ = ∠BPS

    And in the book they've just marked ∠RPQ and ∠BPS down as theta. Is this right???
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Product gradients perpendicular Date
Product of Missing Digits in a Number Wednesday at 7:17 PM
Problem with a product of 2 remainders (polynomials) Jan 29, 2018
Production model problem Sep 1, 2017
Vector dot product and parallel vectors Aug 24, 2017
Help With A Gradient Problem Oct 14, 2013