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Timtime Group Title

Triangle PQR has vertices P(0, 1), Q(0, -4), and R(2, 5). After rotating triangle PQR counterclockwise about the origin 45º, the coordinates of P' to the nearest hundredth are (-0.71, ?

  • one year ago
  • one year ago

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  1. JakeV8 Group Title
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    You don't even have to rotate the triangle... just rotate the point P.

    • one year ago
  2. AbhimanyuPudi Group Title
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    y coordinate is also the same..

    • one year ago
  3. Timtime Group Title
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    so you would basically switch the x points to y points?

    • one year ago
  4. JakeV8 Group Title
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    |dw:1349722508665:dw|

    • one year ago
  5. JakeV8 Group Title
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    no, you can't swap coordinates if you only rotate 45 degrees.

    • one year ago
  6. JakeV8 Group Title
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    The point P is 1 unit away from the origin, right? One unit up on the y axis. The rotated point P' also needs to be 1 unit away from the origin.

    • one year ago
  7. JakeV8 Group Title
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    However, it will be in the middle of that top left quadrant, since a 45 degree rotation doesn't get it all the way down to the x axis

    • one year ago
  8. AbhimanyuPudi Group Title
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    You are doing good dude @JakeV8 I'll leave it to you..

    • one year ago
  9. JakeV8 Group Title
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    @Timtime does that help you get the idea? What I would do next is sketch a line from the origin at that 45 degree angle, and label its length as "1", and then realize that it is a hypotenuse

    • one year ago
  10. JakeV8 Group Title
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    That angled side from P' to the origin makes the hypotenuse of a right triangle if you drop a side down from point P' to the x axis...

    • one year ago
  11. JakeV8 Group Title
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    |dw:1349722859375:dw|

    • one year ago
  12. Timtime Group Title
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    were taught to use use matrices to solve the problem {0 -1} {1 0}

    • one year ago
  13. JakeV8 Group Title
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    interesting. Is a 45 degree rotation something you can express as matrix transformation?

    • one year ago
  14. Timtime Group Title
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    yes its a matrix rotation

    • one year ago
  15. JakeV8 Group Title
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    I know the answer based on the shape and right triangle math, but you could probably generalize the idea of a rotation of any angle. I hadn't thought of that before :)

    • one year ago
  16. Timtime Group Title
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    its weird how my online class is doing it

    • one year ago
  17. JakeV8 Group Title
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    Does knowing the answer help you figure out how to set up the matrix rotation, working backward?

    • one year ago
  18. Timtime Group Title
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    it really would like i have already done like so much with this online working backwords would be different

    • one year ago
  19. JakeV8 Group Title
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    ordinarily I wouldn't just want to give you the answer, but in this case, knowing the answer might help figure out what sort of matrix rotation would have been appropriate. I'm not sure I know how to do it that way though... The point P needs to rotate from (0,1) to (-(sqrt(2)/2, sqrt(2)/2) ), so about (-0.707, 0.707)

    • one year ago
  20. JakeV8 Group Title
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    or (-0.71, 0.71) to the nearest hundredth.

    • one year ago
  21. Timtime Group Title
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    thanks you

    • one year ago
  22. JakeV8 Group Title
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    hope it helps :) good luck!

    • one year ago
  23. afranklin12 Group Title
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    ur cute

    • one year ago
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