1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

A transformation question? HELP PLS

  1. Apr 6, 2006 #1
    a transformation question??? HELP PLS

    Hi, I have this question from classwork and I can't quite figure it out. I know its very simple, but I can't quite understand it....

    If two parallel lines are located ten inches apart, a preimage that is reflected twice through those lines will be _________ inches away from the final image.

    ** I know the answer is twenty, but I'm not sure how/why.
    Could someone help in explaining? It would be very appreciated!!

  2. jcsd
  3. Apr 7, 2006 #2
    check out the attached file. i hope it will be clear from the picture. if not then feel free to tell me. you can also pm me or mail me at murshid_islam@yahoo.com.

    Attached Files:

  4. Apr 7, 2006 #3


    User Avatar
    Science Advisor

    "Reflected twice through those lines"? I will assume that that means "reflected in each of the lines".

    Imagine a point x inches from the first line (for simplicity, take x< 10 and the point is NOT between the two lines). After one reflection, the (image of the) point will be on the other side of the line (and so between the two lines) at distance x inches from it. What distance will it be from the second line now? Call that distance y. After the second reflection, the point will be on the other side of THAT line and the same distance from it. Okay, the total distance will be the original x inches plus the x inches on the other side, plus the y inches the first image was from the second line plus the y inches it was reflected to. What do all those add to?

    Now see if you can do it assuming the original point is between the two lines or if it is more than 10 inches from the first line.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook