Which rule describes a composition of transformations that maps preimage [tex]\( PORS \)[/tex] to image [tex]\( P'ORS' \)[/tex]?

[tex]\[ R_0, 270^\circ \circ T_{-2}(x, y) = \left(-y - 2, x - 2\right) \][/tex]



Answer :

To determine the rule that describes the composition of transformations mapping the pre-image point to the image point, we need to understand the sequence of transformations involved. Here's a step-by-step explanation of each transformation and the resulting coordinates.

1. Translation: The first transformation given is [tex]\( T_{-2}(x, y) \)[/tex]. Translation involves shifting all points by a certain vector. Here, we are shifting each point by [tex]\(-2\)[/tex] units. This means:
[tex]\[ (x, y) \rightarrow (x - 2, y - 2) \][/tex]

2. Rotation: The second transformation is [tex]\( R_{0, 270^\circ} \)[/tex], which represents a rotation about the origin by [tex]\( 270^\circ \)[/tex]. A [tex]\( 270^\circ \)[/tex] rotation counterclockwise can be seen as a [tex]\( 90^\circ \)[/tex] clockwise rotation. This rotation changes the coordinates as follows:
[tex]\[ (x, y) \rightarrow (y, -x) \][/tex]

Let's combine these transformations into a single operation:

1. Apply the translation [tex]\( (x, y) \rightarrow (x - 2, y - 2) \)[/tex]:
After translation, the coordinates will be:
[tex]\[ (x', y') = (x - 2, y - 2) \][/tex]

2. Apply the rotation [tex]\( (y', -x') \)[/tex]:
After the translation, rotate the new coordinates:
[tex]\[ (x', y') \rightarrow (y' - x', -y') = (y - 2, - (x - 2)) = (y - 2, -x + 2) \][/tex]

Therefore, the final transformation that maps the pre-image point to the image point is:
[tex]\[ (x, y) \rightarrow (y - 2, -x + 2) \][/tex]

This complete transformation can be seen as a composition of a translation followed by a rotation.