Use coordinate notation to enter the rule that maps each preimage to its image. Then ic
transformation and confirm that it preserves length and angle measure.
A(-5,5) ► A'(5,5)
B(-2,2) B'(2, 2)
C(-3,2) → C'(2,3)
The transformation is a rotation of ?
(x,y) →
° clockwise about the origin given by the rules
6
5

The transformation is a rotation of 270 degrees clockwise
The transformation preserves length because the length of the image and preimage are the same
The transformation preserves angles because the angles of the image and preimage are the same

Given that:

[tex]A = (-5,5) \to A' = (5,5)[/tex]

[tex]B = (-2,2) \to B' = (2, 2)[/tex]

[tex]C = (-3,2) \to C' = (2,3)[/tex]

By observing the pattern of transformation, the rule is:

[tex](x,y) \to (y,-x)[/tex]

Hence, the transformation is a rotation of 270 degrees clockwise

The length is calculated using the following distance formula:

[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2[/tex]

So, we have:

[tex]AB = \sqrt{(-5 - -2)^2 + (5 - 2)^2} =\sqrt{18[/tex]

[tex]BC = \sqrt{(-2 - -3)^2 + (2 - 2)^2} =1[/tex]

[tex]AC = \sqrt{(-5 - -3)^2 + (5 - 2)^2} =\sqrt{13[/tex]

And

[tex]A'B' = \sqrt{(5 -2)^2 + (5 - 2)^2} =\sqrt{18[/tex]

[tex]B'C' = \sqrt{(2 - 2)^2 + (2 - 3)^2} =1[/tex]

[tex]A'C' = \sqrt{(5 - 2)^2 + (5 - 3)^2} =\sqrt{13[/tex]

By comparing the lengths of the image and the preimage, we can conclude that the transformation preserves length.

The measure of the angles is calculated as follows:

[tex]a^2 = b^2 + c^2 -2ab \cos A[/tex]

So, we have:

[tex]18 = 1 + 13 -2 \times 1 \times \sqrt{13} \cos C[/tex]

[tex]18 - 1 - 13= -2 \times 1 \times \sqrt{13} \cos C[/tex]

[tex]4= -2 \times 1 \times \sqrt{13} \cos C[/tex]

[tex]-2= \sqrt{13} \cos C[/tex]

Make cos C the subject

[tex]\cos C = -0.5547[/tex]

[tex]C = cos^{-1}(-0.5547)[/tex]

[tex]C = 124^o[/tex]

Also, we have:

[tex]\frac{a}{\sin A} =\frac{c}{\sin C}[/tex]

So, we have:

[tex]\frac{1}{\sin A} =\frac{\sqrt{18}}{\sin( 124)}[/tex]

[tex]\frac{1}{\sin A} =5.1175[/tex]

Rewrite as:

[tex]\sin A = \frac{1}{5.1175}[/tex]

[tex]\sin A = 0.1954[/tex]

[tex]A = \sin^{-1}(0.1954)[/tex]

[tex]A = 11^o[/tex]

Also, we have:

[tex]A + B + C = 180^o[/tex] --- sum of angles in a triangle

[tex]11 + B + 124 = 180^o[/tex]

Collect like terms

[tex]B = 180 -11 - 124[/tex]

[tex]B = 45[/tex]

Hence:

[tex]\angle A = 11[/tex] [tex]\angle B = 45[/tex] and [tex]\angle C = 124[/tex]

Read more about coordinate geometry at:

https://brainly.com/question/1601567