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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

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

    • 8 months ago
  2. OpenSessame
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    Those are the reflections

    • 8 months ago
  3. e.mccormick
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    A reflection is usually over a line.... so I want to make sure that is the notation you are using.

    • 8 months ago
  4. OpenSessame
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    I think so...

    • 8 months ago
  5. e.mccormick
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    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.

    • 8 months ago
  6. OpenSessame
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    Okay, but i dont understand what i have to do...

    • 8 months ago
  7. e.mccormick
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    Well, it needs to be an equal distance from the line on the other side of the line.

    • 8 months ago
  8. OpenSessame
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    Where did the line come from????

    • 8 months ago
  9. OpenSessame
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    You lost me..

    • 8 months ago
  10. e.mccormick
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    Ry=-5 \(\leftarrow\) that is the line part.

    • 8 months ago
  11. e.mccormick
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    Let me show you an example.

    • 8 months ago
  12. OpenSessame
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    Okay...Just show me with this one

    • 8 months ago
  13. e.mccormick
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    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.

    • 8 months ago
  14. e.mccormick
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    |dw:1376455572591:dw|

    • 8 months ago
  15. OpenSessame
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    Well, im really lost man...So i wouldnt even get it

    • 8 months ago
  16. OpenSessame
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    That is an example already, im just doing a text review for before i take it. so it doesnt count

    • 8 months ago
  17. e.mccormick
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    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.

    • 8 months ago
  18. e.mccormick
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    |dw:1376455677255:dw|

    • 8 months ago
  19. OpenSessame
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    Do the distance is the number in the equation?

    • 8 months ago
  20. OpenSessame
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    Let me try to do mine...one sec

    • 8 months ago
  21. e.mccormick
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    The distance is the measure from your point and the line.

    • 8 months ago
  22. e.mccormick
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    Yours is done twice.... so see what you get for the first step and I can check it. Then the second step.

    • 8 months ago
  23. OpenSessame
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    |dw:1376455841757:dw|

    • 8 months ago
  24. e.mccormick
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    OK, that is a good (-2,-1).

    • 8 months ago
  25. OpenSessame
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    |dw:1376455914360:dw|

    • 8 months ago
  26. OpenSessame
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    Now what do i do exactly?

    • 8 months ago
  27. e.mccormick
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    OK, so what is the measure of the distance between your dot and the line? |dw:1376456012468:dw|

    • 8 months ago
  28. OpenSessame
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    -3?

    • 8 months ago
  29. e.mccormick
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    Not quite. Tewo problems. This is change in y, not x, and distance is always positive.

    • 8 months ago
  30. OpenSessame
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    so 4?

    • 8 months ago
  31. e.mccormick
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    Yes. So your point is 4 above the line. That means the reflection must be 4 BELOW the line!

    • 8 months ago
  32. OpenSessame
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    so at -5?

    • 8 months ago
  33. OpenSessame
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    wait no....-4?

    • 8 months ago
  34. e.mccormick
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    The line is at -5. So 4 below -5.... or -5-4=?

    • 8 months ago
  35. e.mccormick
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    Above is +, below is -. Similarly, to the left of is -, to the right of is +. Has to do with the Cartesian (xy) plane rules.

    • 8 months ago
  36. OpenSessame
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    Ohhhh so -9?

    • 8 months ago
  37. e.mccormick
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    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|

    • 8 months ago
  38. e.mccormick
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    Now you have the line Rx=1 to do the second reflection over.

    • 8 months ago
  39. OpenSessame
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    okay:) i get it!

    • 8 months ago
  40. OpenSessame
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    |dw:1376456678313:dw|

    • 8 months ago
  41. OpenSessame
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    |dw:1376456695550:dw|

    • 8 months ago
  42. e.mccormick
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    So, how far this time?

    • 8 months ago
  43. OpenSessame
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    3?

    • 8 months ago
  44. e.mccormick
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    \(\large\ddot\smile\)

    • 8 months ago
  45. OpenSessame
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    so the point would me three to the right?

    • 8 months ago
  46. OpenSessame
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    so (4,-9)????

    • 8 months ago
  47. e.mccormick
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    Yes, 3 to the right of 1. And yep, that is it. (4,-9).

    • 8 months ago
  48. OpenSessame
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    wait im gonna see if it matches...

    • 8 months ago
  49. e.mccormick
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    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!

    • 8 months ago
  50. OpenSessame
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    Okay, the textbook just didnt really explain it

    • 8 months ago
  51. e.mccormick
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    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!

    • 8 months ago
  52. OpenSessame
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    yea, thats the right answer! thanks:)

    • 8 months ago
  53. OpenSessame
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    Can you help with some other questions that i dont get?

    • 8 months ago
  54. e.mccormick
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    Probably. I can explain it the same basic way I always do and hopefully it works like this and you get it!

    • 8 months ago
  55. OpenSessame
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    Okay, thanks:)

    • 8 months ago
  56. OpenSessame
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    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

    • 8 months ago
  57. e.mccormick
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    OK. Dilation is going to be the size change. The rigit motion will get it from one place to the other.

    • 8 months ago
  58. OpenSessame
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    Okay...

    • 8 months ago
  59. e.mccormick
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    It is probably easiest to do the dilation first. That will be some multiplication number. It is a ratio of the sides.

    • 8 months ago
  60. OpenSessame
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    Okay...

    • 8 months ago
  61. e.mccormick
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    For example, this would be a dialation by 3..... |dw:1376457753408:dw| So start by finding how long the sides of both rectangles are.

    • 8 months ago
  62. OpenSessame
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    wait is the answer D2*T<0,-6>

    • 8 months ago
  63. e.mccormick
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    2x the size, so yah, a dialation of 2. I didn't do the rest yet, but that would probably do it.

    • 8 months ago
  64. OpenSessame
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    Really??? woooohoooo

    • 8 months ago
  65. e.mccormick
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    Hmmm.... -6.... is that enough on the translation?

    • 8 months ago
  66. OpenSessame
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    Idk...

    • 8 months ago
  67. e.mccormick
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    Well, lets look at this real quick: |dw:1376458028692:dw| Now you need to double the size.

    • 8 months ago
  68. OpenSessame
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    The dialation is times 2?

    • 8 months ago
  69. e.mccormick
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    |dw:1376458060944:dw| OK, that makes them reflections of each other.

    • 8 months ago
  70. e.mccormick
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    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|

    • 8 months ago
  71. OpenSessame
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    so its shifted down -6

    • 8 months ago
  72. e.mccormick
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    Well, is that 6? That bottom corner is at -9....

    • 8 months ago
  73. OpenSessame
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    -10?

    • 8 months ago
  74. e.mccormick
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    or -10, actually.... counted wrong.

    • 8 months ago
  75. e.mccormick
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    So it has to go from 2 above to -10 below... so....

    • 8 months ago
  76. OpenSessame
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    so uhm...It would be 11?

    • 8 months ago
  77. OpenSessame
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    12

    • 8 months ago
  78. e.mccormick
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    Yep! So 2D, T\(\langle 0,-12\rangle\) is more like it.

    • 8 months ago
  79. OpenSessame
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    that isnt one of my options

    • 8 months ago
  80. e.mccormick
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    Hmmm.... AH.... OK... the dialaion is probably also changing the translation! \(2\langle 0,-6\rangle = \langle 0,-12\rangle\)

    • 8 months ago
  81. e.mccormick
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    So my bad. Looks like you did have it right.

    • 8 months ago
  82. OpenSessame
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    YUS:) thanks though!

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

    • 8 months ago
  84. e.mccormick
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    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....

    • 8 months ago
  85. OpenSessame
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    okay!

    • 8 months ago
  86. e.mccormick
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    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?

    • 8 months ago
  87. OpenSessame
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    240 degrees

    • 8 months ago
  88. OpenSessame
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    Thats really all i understand...

    • 8 months ago
  89. e.mccormick
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    r is usually a radius.... but there is no distance....

    • 8 months ago
  90. OpenSessame
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    I think r is the point?

    • 8 months ago
  91. e.mccormick
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    Is it R and not r?

    • 8 months ago
  92. OpenSessame
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    Thats the problem..

    • 8 months ago
  93. e.mccormick
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    hmmm..... Well, 240/30=8 segments.

    • 8 months ago
  94. OpenSessame
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    so 8 to the left or right?

    • 8 months ago
  95. e.mccormick
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    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.

    • 8 months ago
  96. OpenSessame
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    So would it be K???

    • 8 months ago
  97. e.mccormick
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    I would only be guessing without the book and chapter.

    • 8 months ago
  98. OpenSessame
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    Okay...Well for me thats better then nothing!

    • 8 months ago
  99. OpenSessame
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    The vertices of a triangle are P(–8, 6), Q(1, –3), and R(–6, –3). Name the vertices of \[R _{y=x}(PQR)\]

    • 8 months ago
  100. e.mccormick
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    Is that a reflection over the y axis?

    • 8 months ago
  101. OpenSessame
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    I believe it is.

    • 8 months ago
  102. e.mccormick
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    AH HA! Found a reference! http://www.regentsprep.org/Regents/math/geometry/GT6/composition.htm

    • 8 months ago
  103. OpenSessame
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    You did??

    • 8 months ago
  104. OpenSessame
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    OOOOO

    • 8 months ago
  105. e.mccormick
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    The composition part does not matter here.... the \(R_y\) part is reflection over the y axis!

    • 8 months ago
  106. OpenSessame
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    So just make it the other sign?

    • 8 months ago
  107. e.mccormick
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    And even better! http://www.regentsprep.org/Regents/math/geometry/GT5/reviewTranformations.htm

    • 8 months ago
  108. OpenSessame
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    Would the first point be (-6,8)

    • 8 months ago
  109. OpenSessame
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    OHHHH!!! MAKES SENSE!

    • 8 months ago
  110. OpenSessame
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    so (6,-8)

    • 8 months ago
  111. e.mccormick
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    Amazing what finding a reference to the symbols can do!

    • 8 months ago
  112. OpenSessame
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    YEA! Thanks so much man!

    • 8 months ago
  113. OpenSessame
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    Write a sequence of rigid motions that maps ab and xy

    • 8 months ago
  114. OpenSessame
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    LAST ONE AND I WILL UNDERSTAND GEOMETRY!

    • 8 months ago
  115. e.mccormick
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    That composition part also confirms what we talked about with the <0,-6> becoming <0,-12>. The SECOND part of the compositon happens first!

    • 8 months ago
  116. OpenSessame
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    the second part is (AB=XY)

    • 8 months ago
  117. e.mccormick
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    Are those seperate calculations for those two?

    • 8 months ago
  118. OpenSessame
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    The top doesnt matter just look at the lines

    • 8 months ago
  119. e.mccormick
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    Yah, well, then that is some sort of what? Rotation, reflection etc?

    • 8 months ago
  120. OpenSessame
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    Reflection?

    • 8 months ago
  121. e.mccormick
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    Remember that A is going to X and B is going to Y..... should be a huge clue.... YES! Some sort of reflection.

    • 8 months ago
  122. e.mccormick
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    So, you just need to find the line they are reflected over.

    • 8 months ago
  123. e.mccormick
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    It would be half way between them.

    • 8 months ago
  124. OpenSessame
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    0?

    • 8 months ago
  125. e.mccormick
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    Are they both the same distance from x=0?

    • 8 months ago
  126. OpenSessame
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    yES...

    • 8 months ago
  127. e.mccormick
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    Did you count the distance from 0 to X and 0 to A?

    • 8 months ago
  128. OpenSessame
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    no...

    • 8 months ago
  129. e.mccormick
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    Take a second look.... they are not the same.

    • 8 months ago
  130. OpenSessame
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    Oh...

    • 8 months ago
  131. e.mccormick
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    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.

    • 8 months ago
  132. OpenSessame
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    okay...

    • 8 months ago
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