Applets below present a pre-image (blue) and its image (very light blue). Is there a rigid transformation (isometry) from our list that maps the pre-image onto the image?[br][br][i]Once you hypothesize which isometry it is, turn it on and adjust the parameters (vectors, angles, centers, lines ...) to demonstrate that it is possible. If impossible, state it so in your answer below.[/i]
1. Did you find such an isometry? What is it?
2. Did you find such an isometry? What is it?
3. Did you find such an isometry? What is it?
4. Did you find such an isometry? What is it?
Whether you decide that cases above can or cannot be described using our isometries, try to use the applet below to find a placement of the pre-image and image so that there is no isometry (from our list) that can be used.[br][br][i]Start by turning [u]either[/u] direct or indirect image on (what's the difference?). Then move it wherever you want and rotate it (by the point) as much as you want. Find a placement of the image so that it could not be described by using isometries we have in our list.[/i]
Did you find an arrangement of image and pre-image that shows that our list of isometries is incomplete?
If you answered [b]Yes[/b], please leave the shapes in such position (in the applet above).[br]If you answered [b]No[/b], elaborate more. Do you think there is no such arragement or perhaps there is, it's just you did not find it?
Isometries in our list can be used to describe any isometry. You will not find such a position of the pre-image and image that would require an isometry not listed here.
In your own words, describe the difference between a "direct" and "indirect" image.
There are several ways to think about it. Try to make sense of these three:[br][br]1. You may refer to what we are physically doing if we are modeling these transformation: Imaginge two congruent shapes on a table. How we need to be moving the image if we want to obtain a direct image? How is it different if we want get an indirect image?[br][br]2. Use the keyword "orientation", making sure to explain what you mean by it. (Recall that we used triangles and labeled their vertices in a certain way to show the orientation).[br][br]3. Yet another way of describing it would be to see if a reflection is involved. This is a more mathematical way of describing it. #1 and #2 are more intuitive.
In this part you will be asked to find important elements of given transformations. Use the applet at the end to experiment.
In the applet at the end, turn on Reflection / Flip. If you were told that the blue is the pre-image and yellow is its image under a reflection, how would you find the line of this reflection?
Hint is in the definition of line reflection or a description of what it does: Line reflection with the line "l" maps all points on l onto themselves and for all points that are not on l, l is the perpendicular bisector of AA' (where A is pre-image and A' its image).[br][br]Does this help you find the line of reflection?
In the applet at the end, turn on Rotation / Turn. If you were told that the blue is the pre-image and yellow is its image under a rotation, how would you find the center of this rotation? How could you find the angle of the rotation?[br]
Pick a point (B) and its image (B') and connect them with a line segment. We know that the center of the rotation will have the same distance from B and B' (make sure you know why). This means that the center of rotation will lie somewhere on the perpendicular bisector of BB'.[br][br]How would you find the exact position of the center of the rotation? (Think of another pair of pre-image and image and corresponding bisector.)[br][br]Once the center C is found, the angle can be shown as the angle ACA'. Another way of visualizing the angle is to draw two lines AB and A'B'. If using the latter method, make sure you know which angle formed by the two lines is the rotation angle.
In the applet below, turn on Translation / Slide. If you were told that the blue is the pre-image and yellow is its image under a translation, how would you find its vector?
This is is quite straightforward: Any vector XX' will be a translation vector. Be careful though, order matters!