OpenStudy (anonymous):

Help please !!!!!

OpenStudy (anonymous):

Graph a triangle (XYZ) and reflect it over the line y=-x to create triangle X'Y'Z'. Part 1: Describe the transformation using words. Part 2: Draw a line segment from point X to the reflecting line, and then draw a line segment from point X' to the reflecting line. What do you notice about the two line segments you drew? Part 3: Do you think you would see the same characteristic if you drew the line segment connecting Y with the reflecting line and then Y' with the reflecting line? How do you know?

OpenStudy (anonymous):

@Preetha

OpenStudy (anonymous):

@Ashleyisakitty can you help me please?

OpenStudy (anonymous):

@Rina.r

OpenStudy (anonymous):

Triangle XYZ is located at X (-2, 1), Y (-4, -3), and Z (0, -2).

OpenStudy (anonymous):

Sorry

OpenStudy (anonymous):

@Hero @mathslover anyone please ??

OpenStudy (anonymous):

I dont understand parts 2 and 3.

OpenStudy (anonymous):

@ehuman

OpenStudy (anonymous):

Anyone ....?

OpenStudy (anonymous):

I graphed it and did part 1 now as i connect the point to the line of reflection i am confused

OpenStudy (stormswan):

|dw:1387784356740:dw| so pretend this is what you drew. and now you have to reflect this same thing on the other side..

OpenStudy (stormswan):

|dw:1387784491971:dw| so the black part is what you have to do

OpenStudy (stormswan):

but make sure how ever many squares it is on one side, thats how it is on the other.

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

wait but you had to reflect the triangle over y=x

OpenStudy (anonymous):

then draw a segment from X to the line of reflection and X" to the line of reflection

OpenStudy (stormswan):

|dw:1387784703117:dw| right. so it would be this way...

OpenStudy (stormswan):

|dw:1387784771136:dw| and then copy it over on this side.

OpenStudy (stormswan):

btw. this is just an example. these are not the right coordinates. but it is how you are supposed to reflect.

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

I came up with this . Part 1 They both lie on the line of reflection. The triangles overlap each other . They y and x - coordinates were switched. Part 2 Both segments end at the same point in the line of reflection. Part 3 When reflecting an object over the line y=x the x and y intercepts are switched but still congruent so , the distance will be the same from the point to the line of reflection . Is it correct?

OpenStudy (stormswan):

yes :) well done! is that all you needed?

OpenStudy (anonymous):

can you help me with one more question ?

OpenStudy (anonymous):

thanks !:)

OpenStudy (anonymous):

OpenStudy (anonymous):

Determine if the two figures are congruent and explain your answer. If they are congruent, tell which rigid motions were used.

OpenStudy (anonymous):

Can you please help if you can .

OpenStudy (anonymous):

@Stormswan

OpenStudy (stormswan):

well, they are congruent. the word congruent means the same. so if you look at the two figures, they look very much alike. It would be a glide reflection :)

OpenStudy (stormswan):

would that be it?

OpenStudy (anonymous):

i know they are congruent but i really dont know what a glide reflection is

OpenStudy (anonymous):

My teacher wants us to be specific :)

OpenStudy (anonymous):

If its reflection the line of reflection , translation , rule , rotation , angle

OpenStudy (stormswan):

go here :) it has all of the rigid motions. https://www.math.ku.edu/~jmartin/courses/math105-F11/Lectures/chapter11-part3.pdf

OpenStudy (anonymous):

And is it more than one motion

OpenStudy (anonymous):

is the reflection on the y or x axis

OpenStudy (stormswan):

y axis :)

OpenStudy (anonymous):

Thank you so much !! I really appreciated your help ! :)

OpenStudy (stormswan):

no problem :)

OpenStudy (stormswan):

tag me if you need anything else.

OpenStudy (anonymous):

Sure