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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

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 A55 class=

Sagot :

MrRoyal

Coordinate geometry is the use of a 2D plane to represent points.

  • 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

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