Geometry question I need help with this one

The distance between points A and B is 2•√2 which is different from the distance between the image A' and B' which indicates that the transformation is a non rigid transformation.

How can the images and the type of transformation be found?

First part;

The given points are;

A(3, 5), B(5, 3), C(2, 2)

The given transformation is presented as follows;

(x, y) → (3•x, 2•y)

Therefore, the images are;

A(3, 5) → A'(3 × 3, 2 × 5) = A'(9, 10)

B(5, 3) → B'(3 × 5, 2 × 3) = B'(15, 6)

C(2, 2) → C'(3 × 2, 2 × 2) = C'(6, 4)

A(3, 5) → A'(9, 10)
B(5, 3) → B'(15, 6)
C(2, 2) → C'(6, 4)

Second part;

Length of AB = √((5-3)² + (3-5)²) = 2•√2

Length of A'B' = √((15-9)² + (6-10)²) = 2•√13

2•√2 ≠ 2•√13
AB ≠ A'B'

Given that the length of AB ≠ the length of A'B' then we have;

The transformation is not a rigid transformation.

Learn more about rigid transformations here:

https://brainly.com/question/1462871

#SPJ1

MrRoyal MrRoyal · Answer 2 · 2022-07-25T13:22:54+02:00

The points of the image are (9, 10), (15, 6) and (6, 4) and the resulting image is a nonrigid motion.

How to determine the image of the points?

The points are given as:

A = (3, 5)

B = (5, 3)

C = (2, 2)

The transformation is given as:

(x, y) ⇒ (3x, 2y)

So, we have:

A' = (9, 10)

B' = (15, 6)

C' = (6, 4)

Hence, the image of the transformation is (9, 10), (15, 6) and (6, 4)

See attachment for the image

Also, the resulting image is a nonrigid motion.

This is so because the lengths of the image and the preimage are not equal