Question

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.

Answer 1

@SourMunchkin7806

Answer 2

ok give me a second

Answer 3

can you explain to me what an isometric transformation is

Answer 4

Yes... Its when the image is the same size and shape as its preimage

Answer 5

ok well i cant just give you the answer...do me a favor and map these out...once you map them out on a grid you can answer it and ill check

Answer 6

Can I make this on desmos?? I tried earlier and since theres the ' it wasn't wanting to graph it. <-<

Answer 7

yea you can do it on desmos

Answer 8

https://www.desmos.com/calculator/v1q2n1akyy okay this is what I got... if i did it correctly..

Answer 9

do you know how to make them lined...like connect the dots?

Answer 10

No.. how do i do that on here?? i just now started using desmos. so I dont know the ins and outs <-<

Answer 11

ok hang on

Answer 12

https://www.desmos.com/calculator/v1q2n1akyy ok there are the lines now

Answer 13

you can probably paste that link as your "work"...but tell me are they isometric

Answer 14

its still showing me the dots o_o

Answer 15

try this one https://www.desmos.com/calculator/eqiwywdnh4

Answer 16

Okay that worked!

Answer 17

ok so are they isometric

Answer 18

I think so.. They have the same shape..

Answer 19

ok good so you can use that graph as the work just copy and past the second link i sent

Answer 20

can you do the translation...just pick a point either a B or C...and then count how many units to the right and up it is for it to fall back on its second location

Answer 21

Okay.. Not trying to sound stupid.. just wanting to make sure i get it right because I just now started doing this type of stuff...How would I count it up and to the right? like what point would i start from?

Answer 22

ok lets start from A...so at the intersections...where a vertical and horizontal line cross...how many units right and up is A'

Answer 23

ill give it to you and explain if you want

Answer 24

Isometric Transformation is where the shape is either reflected, rotated, or translated without changing the shape or size of the image.

To prove this is only Isometric transformation we would identify the distance between point A and B, and the distance of A' and B'. If the distances are the same then this IS a isometric transformation.

Answer 25

Okay that would be great @SourMunchkin7806 just explain as you go if you don't mind... Sorry I lost internet x_x

Answer 26

no prob so it would be four units right and 4 units up...so just count the boxes over and up till you are at the point and you have the answer

Answer 27

@SourMunchkin7806 that is for B correct?

Answer 28

yes

Answer 29

you dont have to tag me im on the questions

Answer 30

sorry it's a habit

Answer 31

no prob

Answer 32

Okay. I got that part.. then would I do the same with C???

Answer 33

there is no c

Answer 34

Your graph for A is correct Sour, but it says she needs to prove with distance formula...

Answer 35

oh...yea i cant help much with that sorry i dont know it

Answer 36

Sorry I mean from (-5,3) to (-1,7).. And yes I do need to prove with distance formula or my teacher will mark it wrong.. Shes not the nicest person in the world <-<

Answer 37

can he help

Answer 38

Part A:

Isometric Transformation is where the shape is either reflected, rotated, or translated without changing the shape or size of the image.

To prove this is only Isometric transformation we would identify the distance between point A and B, and the distance of A' and B'. If the distances are the same then this IS a isometric transformation.

~~~

So let's plug Point A and Point B into distance formula.

\(\large\bf{Point~A~to~Point~B:\sqrt{(0-0)^{2}+(-1-(-6))^{2}}}\)

Let's do the same for Point A' and Point B'.

\(\large\bf{Point~A'~to~Point~B':\sqrt{4-4)^{2}+(3-(-2))^{2}}}\)

The outcome needs to be the same, if it is then it is isometric.

Answer 39

I'm sorry not trying to sound stupid.. What program can I use to find the distance. Sorry for a late reply. Its like I told shadow, my internet provider was bought out so my internets been going in and out badly.

Answer 40

Uhhh you can't use a program, the equation simply needs to be simplified....you can use mathway.com but it would be best to do it by hand....

Answer 41

√(4-4)²+(3-(-2))=5
√(0-0)²+(-1-(-6))²=6
sorry. I forgot I had to simplify :')

Answer 42

So no, they aren't isometric.

Answer 43

You there @563blackghost ?

Answer 44

the simplifying of the second one is incorrect.

Answer 45

Then what would it be? That's what I'm coming up with. Not too sure where I went wrong on it. ._.

Answer 46

You went wrong with the subtraction of -1 and 6.

\(\large\bf{-1-(-6) \rightarrow -1 + 6 = 5}\)

The squares cancel so our distance is 5.

So the two are Isometric.

Answer 47

Oops. My bad.. I'm sorry ._.

Answer 48

that's fine, now you know ^-^