In geometry, transformations are used to move points from one position to another.
- The transformation is a rigid transformation given by [tex](x,y) \to (x + 1, y-2)[/tex].
- It preserves lengths and angles
The preimage is given as:
[tex]S =(-3,4)[/tex]
[tex]T = (3,4)[/tex]
[tex]U = (-1,0)[/tex]
The image is given as:
[tex]S' = (-2,2)[/tex]
[tex]T' =(4,2)[/tex]
[tex]U' = (0,-2)[/tex]
Using points S and S' as our reference point, we have:
[tex]S =(-3,4)[/tex] and [tex]S' = (-2,2)[/tex]
Calculate transformation
[tex](x,y) \to S -S'[/tex]
[tex](x,y) \to (-2,2)-(-3,4)[/tex]
Rewrite as:
[tex](x,y) \to (-2--3,2-4)[/tex]
[tex](x,y) \to (-2+3,2-4)[/tex]
[tex](x,y) \to (1,-2)[/tex]
So, the transformation from the preimage to image is:
[tex](x,y) \to (x + 1, y-2)[/tex]
Calculate the lengths of the sides of the preimage and the image
Using the following distance formula, we have:
[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]
So, we have:
[tex]ST = \sqrt{(-3 - 3)^2 + (4 - 4)^2} = 6[/tex]
[tex]SU = \sqrt{(-3 - -1)^2 + (4 - 0)^2} = \sqrt{20[/tex]
[tex]TU = \sqrt{(3 - -1)^2 + (4 - 0)^2} = \sqrt{32[/tex]
and
[tex]S'T' = \sqrt{(-2 - 4)^2 + (2 - 2)^2} = 6[/tex]
[tex]S'U' = \sqrt{(-2 - 0)^2 + (2 -- 2)^2} = \sqrt{20[/tex]
[tex]T'U' = \sqrt{(4 -0)^2 + (2 -- 2)^2} = \sqrt{32[/tex]
Since
[tex]ST = S'T'= 6[/tex]
[tex]SU = S'U'= \sqrt{20[/tex]
and
[tex]TU = T'U'= \sqrt{32[/tex]
Then, the transformation preserves length
From the graph of [tex]\triangle STU[/tex] and [tex]\triangle S'T'U'[/tex]
Since
[tex]\angle S = \angle S' = 76^o[/tex]
[tex]\angle T = \angle T' = 45^o[/tex]
[tex]\angle U = \angle U' = 59^o[/tex]
Then, the transformation preserves angle measure.
Read more about transformation at:
https://brainly.com/question/12865301
