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Reflection of triangle over a line

  1. Apr 18, 2017 #1
    1. The problem statement, all variables and given/known data
    How to rotate a triangle over a line? I have provided a attachment for it?

    2. Relevant equations

    Y= mx +c
    I dont know the eq for reflection
    3. The attempt at a solution
    My plan is find the eq of line. I have got points for the end points of the line but i dont know how to reflect the triangle over the line which is not straight. I searched the internet but i cant find any eq. Eq are provided for reflecting over x-axis, y-axis , over the line x=2 & over the line x=y but no equations are shown for rotating object over an oblique line. One site gave the method to find the eq of line which was used to rotate the triangle but it dont give any example for reverse case. Its link was:

    http://www.rcboe.org/cms/lib010/GA01903614/Centricity/Domain/1030/Reflections.pdf
    I have attached the figure. I want to rotate the triangle over the dark line.
    Some body please guide me.

    Zulfi.
     

    Attached Files:

  2. jcsd
  3. Apr 18, 2017 #2

    BvU

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    Hello Zak,

    Reflection of triangle over a line

    First thing to do is find the right question to ask.
    So far, title, link and missing equation are about reflection. The figure title says reflection too. But I see a 6 point star and a fat line. The fat line connects two tips but has little to do with any reflection nor with any rotation.

    So: make up your mind :smile:. What is it really that you want to do ?

    I also note your vocabulary differs from that in the link (which is directed at teachers, by the way). Handy to have the answers filled in already, I must say. Do you understand it all ?
     
  4. Apr 18, 2017 #3
    Mod note: Emphasis added to highlight changes from original post.
    Hi,
    Thanks for your response. I would make some corrections:
    <<
    Eq are provided for reflecting over x-axis, y-axis , over the line x=2 & over the line x=y but no equations are shown for rotating object over an oblique line.
    >>
    Correction:
    Equations are provided for reflecting over x-axis, y-axis , over the line x=2 & over the line x=y but no equations found for reflecting object over an oblique line.

    If this is not clear please let me know.
    <<
    One site gave the method to find the eq of line which was used to rotate the triangle but it dont give any example for reverse case.
    >>
    Correction:

    One site gave the method to find the equation of line which produced the reflection of a triangle over that line but it did not give any example for reverse case i.e. for reflecting the triangle over a line (Note at same places its written reflection of triangle across a line. I don't know if across or over the line same thing?).

    <<
    I want to rotate the triangle over the dark line.>>
    I want to do the reflection of triangle over the dark line.
    <<
    Do you understand it all ?
    >>
    I did not try to solve any example because I was looking for the solution of my problem. At first glance some problems looked understandable.

    Kindly guide me.

    Zulfi.
     
    Last edited by a moderator: Apr 18, 2017
  5. Apr 18, 2017 #4

    Mark44

    Staff: Mentor

    Thread reopened.

    If a point is reflected across a line, a line segment joining the new (reflected) point and the old point will be perpendicular to the line of reflection, and both points will be the same distance from the line of reflection, and on either side of it.

    If a figure is rotated, the pole or point about which rotation occurs has to be stated. For example, if the line segment between (1, 1) and (3, 3) is rotated 45° clockwise around the origin, the rotated segment will lie along the x-axis between ##(\sqrt 2, 0)## and ##(3\sqrt 2, 0)##. Both line segments have the same length, ##2\sqrt 2##, but their orientation is different.
    .
     
    Last edited: Apr 18, 2017
  6. Apr 18, 2017 #5

    BvU

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    Hi zak, instead of using << and >> you could use the tags [quote] and [\quote] or (even easier) select the text in question in the post and click +Quote to copy it and click 'insert quotes' to paste it -- adorned with quote tags.
    Equations are provided for reflecting over x-axis, y-axis , over the line x=2 & over the line x=y but no equations found for
    reflecting object over an oblique line.


    It looks (e.g. in your link) as if the customary expression is 'to reflect across a line' -- but as a non-native english speaker I have no authority in this (I think I used ' with respect to a line' thus far in my career).

    Back to the issue.
    and you want to extend that to the line y = mx + c.

    One way to do that is to reduce the latter to one of the known ones using a series of coordinate transformations: shift down by c, rotate around the origin until the mirror line is vertical (i.e. over ##\pi/2 - \alpha\ \ ## where ##\ \ \tan\alpha = m## ), then reflect wrt the y axis (i.e. change x to -x), rotate back to mx ( over ##\ \alpha - \pi/2## ) and shift up by c.

    Something for a computer....(which is why the books stay with the simple cases).

    Now a concern of Mark and me: you mention reflection over a line which is not straight. that would make things different (and difficult). What are you referring to ?
     
  7. Apr 18, 2017 #6

    haruspex

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    Judging from post #1, Zulfi just meant a line which is oblique, i.e. not parallel to an axis.
     
  8. Apr 20, 2017 #7
    Hi,
    Thanks for updating my post. I have made a figure and specified the coordinates of the original and the rotated triangle. This would help to understand my problem.

    coordinates of the rotated superimposed triangle2.jpg

    I want to do reflection of the superimposed triangles over the line formed by the points (x1r, y1r) and (x2, y2).

    I dont know the actual equation to find out the coordinates of (x1reflected, y1reflected) , (x2r_reflected, y2r_reflected), (x3reflected, y3reflected) and (x3r_reflected, y3r_reflected). Somebody please guide me.

    Zulfi.
     
  9. Apr 20, 2017 #8

    haruspex

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    Do you know how to find the point on a line that is nearest a given point? From there it is easy.
     
  10. Apr 20, 2017 #9
  11. Apr 20, 2017 #10

    Mark44

    Staff: Mentor

    No.
    To use the distance formula, you need both points. @haruspex asked whether you knew how to find the point on a line that is closest to a given point.
     
  12. Apr 20, 2017 #11
    Hi,
    I found the formula at the following link:

    https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line

    i would try to write it:

    The point on this line which is closest to (x0,y0) has coordinates:[3]

    https://wikimedia.org/api/rest_v1/media/math/render/svg/e272ef09915cac216abc89acba3d0eedea77cf30
    My logic was to use the distance formula because in this case we know the endpoints of the line.So we can find out the distance between the end points & the given point. Using the endpoints we can also find the mid-point of the line. Thus we have 3 distances now.. Any two minimum distances mean that the point of lowest distance may lie between them.

    Kindly reply me.
    Zulfi.
     
  13. Apr 21, 2017 #12

    haruspex

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    The distance is not helpful. Once you have (x0, y0) and the point on the line that is nearest to it, finding the mirror image in the line is trivial.
     
  14. Apr 21, 2017 #13
    Hi,

    Actually i never did this before. Kindly guide me, what would be the equation for the reflected points.

    Zulfi.
     
  15. May 8, 2017 #14
  16. May 8, 2017 #15

    haruspex

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    I am sorry, I did not notice your 21st April reply.
    For what it is worth, if you have a point X=(x,y) and the nearest point to it on the line is X'=(x',y'), the mirror image, X", of the point can be found by considering the vector XX'. This must equal the vector X'X", so
    X"-X'=X'-X
    X'=2X'-X.
     
  17. May 8, 2017 #16

    BvU

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