Question

use coordinate notation to write the rule that maps each preimage to its image. Then confirm that the transformation is not a rigid motion.Number 8

Answer 1

In geometry, transformation involves changing the position and/or size of a shape.

The coordinate notations are:

• $(x,y) \to (\frac{1}{2}x,\frac{1}{2}y)$.
• $(x,y) \to (2x,y)$.

Both transformations are not rigid

$7.\ \triangle ABC \to \triangle A'B'C'$

Using coordinates A and A', as a reference;

We have:

$A = (6,6)$

$A' = (3,3)$

Divide A' by A to get the scale ratio (k)

$k = \frac{A'}{A}$

$k = \frac{(3,3)}{(6,6)}$

Factorize

$k = \frac{(3,3)}{2 \times (3,3)}$

$k = \frac{1}{2}$

Hence, the coordinate notation is:

$(x,y) \to (\frac{1}{2}x,\frac{1}{2}y)$

Calculate distance AB and A'B' using:

$d = \sqrt{(x_2 - x_1)^2 +(y_2 - y_1)^2}$

So, we have:

$AB = \sqrt{(6 - 4)^2 +(6 - -2)^2} = \sqrt{68}$

$A'B' = \sqrt{(3 - 2)^2 +(3 - -1)^2} = \sqrt{17}$

Because

$AB \ne A'B'$

We've confirmed that the transformation is not rigid

$8.\ \triangle FGH \to \triangle F'G'H'$

Using coordinates F and F';

We have:

$F = (-1,1)$

$F' = (-2,1)$

Divide the x-coordinate of F' by F to get the scale ratio (k) because the y-coordinates of the pre-image and the image are the same

$k = \frac{F'}{F}$

$k = \frac{-2}{-1}$

$k = 2$

Hence, the coordinate notation is:

$(x,y) \to (2x,y)$

Calculate distance FG and F'G' using:

$d = \sqrt{(x_2 - x_1)^2 +(y_2 - y_1)^2}$

So, we have:

$FG = \sqrt{(-1 - 1)^2 +(1 - -1)^2} = \sqrt{8}$

$F'G' = \sqrt{(-2 - 2)^2 +(1 - -1)^2} = \sqrt{20}$

Because

$FG \ne F'G'$

We've confirmed that the transformation is not rigid

Read more about transformations at:

https://brainly.com/question/11709244