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Timtime

  • 2 years ago

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, ?

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  1. JakeV8
    • 2 years ago
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    You don't even have to rotate the triangle... just rotate the point P.

  2. AbhimanyuPudi
    • 2 years ago
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    y coordinate is also the same..

  3. Timtime
    • 2 years ago
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    so you would basically switch the x points to y points?

  4. JakeV8
    • 2 years ago
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    |dw:1349722508665:dw|

  5. JakeV8
    • 2 years ago
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    no, you can't swap coordinates if you only rotate 45 degrees.

  6. JakeV8
    • 2 years ago
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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.

  7. JakeV8
    • 2 years ago
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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

  8. AbhimanyuPudi
    • 2 years ago
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    You are doing good dude @JakeV8 I'll leave it to you..

  9. JakeV8
    • 2 years ago
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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

  10. JakeV8
    • 2 years ago
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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...

  11. JakeV8
    • 2 years ago
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    |dw:1349722859375:dw|

  12. Timtime
    • 2 years ago
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    were taught to use use matrices to solve the problem {0 -1} {1 0}

  13. JakeV8
    • 2 years ago
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    interesting. Is a 45 degree rotation something you can express as matrix transformation?

  14. Timtime
    • 2 years ago
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    yes its a matrix rotation

  15. JakeV8
    • 2 years ago
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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 :)

  16. Timtime
    • 2 years ago
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    its weird how my online class is doing it

  17. JakeV8
    • 2 years ago
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    Does knowing the answer help you figure out how to set up the matrix rotation, working backward?

  18. Timtime
    • 2 years ago
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    it really would like i have already done like so much with this online working backwords would be different

  19. JakeV8
    • 2 years ago
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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)

  20. JakeV8
    • 2 years ago
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    or (-0.71, 0.71) to the nearest hundredth.

  21. Timtime
    • 2 years ago
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    thanks you

  22. JakeV8
    • 2 years ago
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    hope it helps :) good luck!

  23. afranklin12
    • 2 years ago
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    ur cute

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