Medal + Fanning for verifying my answer
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Hurry up
Loll
Okay one sec
Hahaha Not Funny Hurry Or I Ain't Helping
Trapezoid ABCD is reflected over the line y = x. What rule shows the input and output of the reflection, and what is the new coordinate of A'?
Shoot
Alright sorry
Feeling Annoyed~
(x,y)→(y,-x); A' is at (1, 5) (x,y)→(y,x); A' is at (1, -5) (x,y)→(-x,y); A' is at (5, 1) (x,y)→(-x,-y); A' is at (5, -1)
Sorry but that's it
>.< Idk This @ganeshie8 and I'm jp people this is my twin brother lol
loll yeah
I think the answer is (x,y)→(-x,y); A' is at (5, 1)
IDK either. It's got me beat.
Loll Ok
@SolomonZelman
I don't feel like going over the comments, if you tagged, tell me your question again please.
Ok Trapezoid ABCD is reflected over the line y = x. What rule shows the input and output of the reflection, and what is the new coordinate of A'?
(x,y)→(y,-x); A' is at (1, 5) (x,y)→(y,x); A' is at (1, -5) (x,y)→(-x,y); A' is at (5, 1) (x,y)→(-x,-y); A' is at (5, -1)
@SolomonZelman
assuming it was in quadrant I it now goes to quadrant 3, (you are reflecting across the diagonal) so from having positive x and positive y i.e. (x,y) YOU NOW GET neg. x and neg. y i.e. (-x,-y) Hope this is good enough :)
so would it be (x,y)→(-x,y); A' is at (5, 1) @SolomonZelman
hehe, I explained it as well as I could, but you still don't get it. that means you don't pay attention in class.
I'm home schooled so when I don't get something right away I asked people on Open study if I understood it then would I have posted the question? Please don't assume that since I don't get it I don't pay attention @SolomonZelman
Alright, sorry. I can self-teach it to you. I'll be asking questions and you reply. Do you know all the quadrants (I , II, III, IV ) ? Yes or No
No I'm taking a pretest so I don't know anything in Unit 2
I don't know what can you possible refer to as "unit 2" but anyway, so you don't know whta the quadrants are, right?
No
Ok... are you familiar with this pic below? |dw:1395370848144:dw|
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