MATH SOLVE

2 months ago

Q:
# Triangle ABC is translated onto its image, triangle A'B'C'. Use the given vertices of the triangles to answer parts A and B. A(-6, 0) B(-1, 0) C(-5, 3) A'(-2, 4) B'(3, 4) C'(-1, 7) Part A: Use the distance formula to prove that the translation was an isometric transformation. Include all of your work in your final answer. Part B: Use complete sentences to describe the translation that maps triangle ABC onto its image.

Accepted Solution

A:

Part A:

An isometric transformation is a type of transformation where the original shape and size of the pre-image is not altered in the image.

To show that the translation was an isometric transformation, we show that the distance between any two points in the pre-image is equal to the distance between the corresponding points in the image.

Consuder, line AB, the distance between point A and point B is given by:

[tex]d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ = \sqrt{(-1-(-6))^2+(0-0)^2} = \sqrt{(-1+6)^2+0^2} \\ \\ = \sqrt{5^2+0} = \sqrt{25} =5[/tex]

The distance between point A' and point B' in the image is given by:

[tex]d= \sqrt{(3-(-2))^2+(4-4)^2} \\ \\ = \sqrt{(3+2)^2+0^2} = \sqrt{5^2+0} \\ \\ = \sqrt{25} =5[/tex]

Similarly checking other points of the pre-image against the corresponding points of the image shows that the size of the pre-image is preserved in the image.

Part 2:

The translation that maps the triangle ABC onto its image are:

Triangle ABC was shifted 4 units to the right.

Triangle ABC was shifted 4 units up.

An isometric transformation is a type of transformation where the original shape and size of the pre-image is not altered in the image.

To show that the translation was an isometric transformation, we show that the distance between any two points in the pre-image is equal to the distance between the corresponding points in the image.

Consuder, line AB, the distance between point A and point B is given by:

[tex]d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ = \sqrt{(-1-(-6))^2+(0-0)^2} = \sqrt{(-1+6)^2+0^2} \\ \\ = \sqrt{5^2+0} = \sqrt{25} =5[/tex]

The distance between point A' and point B' in the image is given by:

[tex]d= \sqrt{(3-(-2))^2+(4-4)^2} \\ \\ = \sqrt{(3+2)^2+0^2} = \sqrt{5^2+0} \\ \\ = \sqrt{25} =5[/tex]

Similarly checking other points of the pre-image against the corresponding points of the image shows that the size of the pre-image is preserved in the image.

Part 2:

The translation that maps the triangle ABC onto its image are:

Triangle ABC was shifted 4 units to the right.

Triangle ABC was shifted 4 units up.