## OpenSessame Find the image of P(–2, –1) after two reflections; first Ry=-5 and then Rx=1. A. (–2, –1) B. (–1, –6) C. (4, –9) D. (1, –5) 8 months ago 8 months ago

1. e.mccormick

So the Ry=-5 and Rx=1 are the lines you need to reflect across?

2. OpenSessame

Those are the reflections

3. e.mccormick

A reflection is usually over a line.... so I want to make sure that is the notation you are using.

4. OpenSessame

I think so...

5. e.mccormick

http://www.mathopenref.com/reflect.html Like it says there, "A transformation where each point in a shape appears at an equal distance on the opposite side of a given line - the line of reflection. " So, start by just doing a rough sketch. Does not need to be exact, but you need to find where the dot and line would be and how far the line is from the dot. If you can do that with math, great! But a sketch can help.

6. OpenSessame

Okay, but i dont understand what i have to do...

7. e.mccormick

Well, it needs to be an equal distance from the line on the other side of the line.

8. OpenSessame

Where did the line come from????

9. OpenSessame

You lost me..

10. e.mccormick

Ry=-5 $$\leftarrow$$ that is the line part.

11. e.mccormick

Let me show you an example.

12. OpenSessame

Okay...Just show me with this one

13. e.mccormick

If I work this that is me giving you an answer. If I work an example you can follow the process and answer your own homework. Lets say I take this random dot: |dw:1376455485961:dw| And lets say I have Ry=1, so the line where every Y is 1.

14. e.mccormick

|dw:1376455572591:dw|

15. OpenSessame

Well, im really lost man...So i wouldnt even get it

16. OpenSessame

That is an example already, im just doing a text review for before i take it. so it doesnt count

17. e.mccormick

The important measure is the distance from the dot to the line. |dw:1376455620753:dw| Because I need to put the reflection the SAME DISTANCE on the other side of the line.

18. e.mccormick

|dw:1376455677255:dw|

19. OpenSessame

Do the distance is the number in the equation?

20. OpenSessame

Let me try to do mine...one sec

21. e.mccormick

The distance is the measure from your point and the line.

22. e.mccormick

Yours is done twice.... so see what you get for the first step and I can check it. Then the second step.

23. OpenSessame

|dw:1376455841757:dw|

24. e.mccormick

OK, that is a good (-2,-1).

25. OpenSessame

|dw:1376455914360:dw|

26. OpenSessame

Now what do i do exactly?

27. e.mccormick

OK, so what is the measure of the distance between your dot and the line? |dw:1376456012468:dw|

28. OpenSessame

-3?

29. e.mccormick

Not quite. Tewo problems. This is change in y, not x, and distance is always positive.

30. OpenSessame

so 4?

31. e.mccormick

Yes. So your point is 4 above the line. That means the reflection must be 4 BELOW the line!

32. OpenSessame

so at -5?

33. OpenSessame

wait no....-4?

34. e.mccormick

The line is at -5. So 4 below -5.... or -5-4=?

35. e.mccormick

Above is +, below is -. Similarly, to the left of is -, to the right of is +. Has to do with the Cartesian (xy) plane rules.

36. OpenSessame

Ohhhh so -9?

37. e.mccormick

Yes. That means the X is the same and the Y is now -9. So, the point after 1 translation is at (-2,-9). |dw:1376456612998:dw|

38. e.mccormick

Now you have the line Rx=1 to do the second reflection over.

39. OpenSessame

okay:) i get it!

40. OpenSessame

|dw:1376456678313:dw|

41. OpenSessame

|dw:1376456695550:dw|

42. e.mccormick

So, how far this time?

43. OpenSessame

3?

44. e.mccormick

$$\large\ddot\smile$$

45. OpenSessame

so the point would me three to the right?

46. OpenSessame

so (4,-9)????

47. e.mccormick

Yes, 3 to the right of 1. And yep, that is it. (4,-9).

48. OpenSessame

wait im gonna see if it matches...

49. e.mccormick

Now, once you get good at the math, you should not need a sketch, but if yuou are ever in doubt, the rough sketch is there to help!

50. OpenSessame

Okay, the textbook just didnt really explain it

51. e.mccormick

The thing I think most people mes up here is the line.... because x=0 is the y axis, and y=0 is the x axis. So they give you a y line but you draw it the same way as the x axis and that can be confusing!

52. OpenSessame

yea, thats the right answer! thanks:)

53. OpenSessame

Can you help with some other questions that i dont get?

54. e.mccormick

Probably. I can explain it the same basic way I always do and hopefully it works like this and you get it!

55. OpenSessame

Okay, thanks:)

56. OpenSessame

What composition of rigid motions and a dilation maps EFGH to the dashed figure? https://study.ashworthcollege.edu/access/content/group/45b8c516-1008-46d7-aa1d-bb9b62c786ff/geometry_exam_9_files/mc020-1.jpg

57. e.mccormick

OK. Dilation is going to be the size change. The rigit motion will get it from one place to the other.

58. OpenSessame

Okay...

59. e.mccormick

It is probably easiest to do the dilation first. That will be some multiplication number. It is a ratio of the sides.

60. OpenSessame

Okay...

61. e.mccormick

For example, this would be a dialation by 3..... |dw:1376457753408:dw| So start by finding how long the sides of both rectangles are.

62. OpenSessame

63. e.mccormick

2x the size, so yah, a dialation of 2. I didn't do the rest yet, but that would probably do it.

64. OpenSessame

Really??? woooohoooo

65. e.mccormick

Hmmm.... -6.... is that enough on the translation?

66. OpenSessame

Idk...

67. e.mccormick

Well, lets look at this real quick: |dw:1376458028692:dw| Now you need to double the size.

68. OpenSessame

The dialation is times 2?

69. e.mccormick

|dw:1376458060944:dw| OK, that makes them reflections of each other.

70. e.mccormick

So now, one of two things needs to happne.... It either needs to be reflected over the x axis, OR, corner H needs to go all the way down to the bottom corner. |dw:1376458143764:dw|

71. OpenSessame

so its shifted down -6

72. e.mccormick

Well, is that 6? That bottom corner is at -9....

73. OpenSessame

-10?

74. e.mccormick

or -10, actually.... counted wrong.

75. e.mccormick

So it has to go from 2 above to -10 below... so....

76. OpenSessame

so uhm...It would be 11?

77. OpenSessame

12

78. e.mccormick

Yep! So 2D, T$$\langle 0,-12\rangle$$ is more like it.

79. OpenSessame

that isnt one of my options

80. e.mccormick

Hmmm.... AH.... OK... the dialaion is probably also changing the translation! $$2\langle 0,-6\rangle = \langle 0,-12\rangle$$

81. e.mccormick

So my bad. Looks like you did have it right.

82. OpenSessame

YUS:) thanks though!

83. OpenSessame

The hexagon GIKMPR and FJN are regular. The dashed line segments form 30° angles. What is r(240drg,0)(G)

84. OpenSessame
85. e.mccormick

Well, the good news is you should understand the why a bit better! By exploring these, it lets you understand the why of it all....

86. OpenSessame

okay!

87. e.mccormick

Hmmm..... I have not seen that form of question.... my geometry was a while back, and they have changed some of the things they do... Do you know what it is looking for, any terms? An angle?

88. OpenSessame

240 degrees

89. OpenSessame

Thats really all i understand...

90. e.mccormick

r is usually a radius.... but there is no distance....

91. OpenSessame

I think r is the point?

92. e.mccormick

Is it R and not r?

93. OpenSessame
94. OpenSessame

Thats the problem..

95. e.mccormick

hmmm..... Well, 240/30=8 segments.

96. OpenSessame

so 8 to the left or right?

97. e.mccormick

Well, if r is right... but that would be a guess on my part. Right 8 of G? I would need to see the reference to know they type of peoblem to get this one.

98. OpenSessame

So would it be K???

99. e.mccormick

I would only be guessing without the book and chapter.

100. OpenSessame

Okay...Well for me thats better then nothing!

101. OpenSessame

The vertices of a triangle are P(–8, 6), Q(1, –3), and R(–6, –3). Name the vertices of $R _{y=x}(PQR)$

102. e.mccormick

Is that a reflection over the y axis?

103. OpenSessame

I believe it is.

104. e.mccormick

AH HA! Found a reference! http://www.regentsprep.org/Regents/math/geometry/GT6/composition.htm

105. OpenSessame

You did??

106. OpenSessame

OOOOO

107. e.mccormick

The composition part does not matter here.... the $$R_y$$ part is reflection over the y axis!

108. OpenSessame

So just make it the other sign?

109. e.mccormick
110. OpenSessame

Would the first point be (-6,8)

111. OpenSessame

OHHHH!!! MAKES SENSE!

112. OpenSessame

so (6,-8)

113. e.mccormick

Amazing what finding a reference to the symbols can do!

114. OpenSessame

YEA! Thanks so much man!

115. OpenSessame

Write a sequence of rigid motions that maps ab and xy

116. OpenSessame
117. OpenSessame

LAST ONE AND I WILL UNDERSTAND GEOMETRY!

118. e.mccormick

That composition part also confirms what we talked about with the <0,-6> becoming <0,-12>. The SECOND part of the compositon happens first!

119. OpenSessame

the second part is (AB=XY)

120. e.mccormick

Are those seperate calculations for those two?

121. OpenSessame

The top doesnt matter just look at the lines

122. e.mccormick

Yah, well, then that is some sort of what? Rotation, reflection etc?

123. OpenSessame

Reflection?

124. e.mccormick

Remember that A is going to X and B is going to Y..... should be a huge clue.... YES! Some sort of reflection.

125. e.mccormick

So, you just need to find the line they are reflected over.

126. e.mccormick

It would be half way between them.

127. OpenSessame

0?

128. e.mccormick

Are they both the same distance from x=0?

129. OpenSessame

yES...

130. e.mccormick

Did you count the distance from 0 to X and 0 to A?

131. OpenSessame

no...

132. e.mccormick

Take a second look.... they are not the same.

133. OpenSessame

Oh...

134. e.mccormick

You want the line half way between. So take the distance between the two points and divide by 2. It will be that far to the left of the right most line, and right of the left most line.

135. OpenSessame

okay...