anonymous
  • anonymous
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, ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
You don't even have to rotate the triangle... just rotate the point P.
anonymous
  • anonymous
y coordinate is also the same..
anonymous
  • anonymous
so you would basically switch the x points to y points?

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anonymous
  • anonymous
|dw:1349722508665:dw|
anonymous
  • anonymous
no, you can't swap coordinates if you only rotate 45 degrees.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
You are doing good dude @JakeV8 I'll leave it to you..
anonymous
  • anonymous
@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
anonymous
  • anonymous
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...
anonymous
  • anonymous
|dw:1349722859375:dw|
anonymous
  • anonymous
were taught to use use matrices to solve the problem {0 -1} {1 0}
anonymous
  • anonymous
interesting. Is a 45 degree rotation something you can express as matrix transformation?
anonymous
  • anonymous
yes its a matrix rotation
anonymous
  • anonymous
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 :)
anonymous
  • anonymous
its weird how my online class is doing it
anonymous
  • anonymous
Does knowing the answer help you figure out how to set up the matrix rotation, working backward?
anonymous
  • anonymous
it really would like i have already done like so much with this online working backwords would be different
anonymous
  • anonymous
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)
anonymous
  • anonymous
or (-0.71, 0.71) to the nearest hundredth.
anonymous
  • anonymous
thanks you
anonymous
  • anonymous
hope it helps :) good luck!
anonymous
  • anonymous
ur cute

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