Yes, all transformations preserve parallelism and collinearity. Explanation: See solution.
Practice makes perfect
We have learned that translations, reflections, and rotations are isometries, so they preserve angle measures and distances. Therefore, these three transformations preserve parallelism and collinearity.
Transformation
Parallelism
Collinearity
Translations
✓
✓
Reflections
✓
✓
Rotations
✓
✓
Glide Reflections
✓
✓
We only need to check whether dilations preserve these properties or not.
Parallelism
Let's consider AB and CD so that AB ∥ CD. Also, we will consider a dilation with center at P and scale factor k.
Next, let's assume that k> 1 and perform the dilation (enlargement) of the segments.
By definition of dilation, we have that PA'=k(PA) and PB'=k(PB). Consequently, we obtain two pairs of sides that are proportional.
PA'/PA = PB'/PB = k
Additionally, notice that ∠ BPA is common for both △ ABP and △ A'B'P. Therefore, by the Side-Angle-Side (SAS) Similarity Theorem we conclude that △ ABP ~ △ A'B'P. This implies that ∠ ABP ≅ ∠ A'B'P.
Since ∠ ABP and ∠ A'B'P are corresponding angles, by the Converse of the Corresponding Angles Theorem, we have that AB ∥ A'B'. Repeating the process for CD and C'D', we will get that CD ∥ C'D'.
This proves that the dilations preserve parallelism. Keep in mind that if the dilation is a reduction, the proof is similar.
Collinearity
Let's consider three collinear points A, B, and C and a dilation with center at P and scale factor k.
In this part, we will assume that k< 1 and perform the dilation (reduction) to the points above.
By definition of dilation, we have that PA' = k(PA), PB' = k(PB), and PC' = k(PC). This gives us three pairs of proportional segments.
PA'/PA = PB'/PB = PC'/PC = k
Additionally, we can see that ∠ A'PB' is common for both △ A'B'P and △ ABP. Then, by the Side-Angle-Side (SAS) Similarity Theorem, we have that △ A'B'P ~ △ ABP.
Due to the similarity of the triangles, we conclude that ∠ A'B'P ≅ ∠ ABP and so, by the Converse of the Corresponding Angles Theorem, we get that A'B' ∥ AB. Applying a similar reasoning, we get also that △ B'C'P ~ △ BCP.
As before, by the Converse of the Corresponding Angles Theorem we get that BC ∥ B'C'. Notice that AB∥ BC because A, B, and C are collinear.
Since A'B' and B'C' are parallel and pass through the same point, we have that A', B', and C' are collinear. In conclusion, collinearity is preserved under dilations too.
