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anonymous
 3 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, ?
anonymous
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You don't even have to rotate the triangle... just rotate the point P.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y coordinate is also the same..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you would basically switch the x points to y points?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349722508665:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no, you can't swap coordinates if you only rotate 45 degrees.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0However, 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
 3 years ago
Best ResponseYou've already chosen the best response.0You are doing good dude @JakeV8 I'll leave it to you..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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
 3 years ago
Best ResponseYou've already chosen the best response.0That 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349722859375:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0were taught to use use matrices to solve the problem {0 1} {1 0}

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0interesting. Is a 45 degree rotation something you can express as matrix transformation?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes its a matrix rotation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0its weird how my online class is doing it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does knowing the answer help you figure out how to set up the matrix rotation, working backward?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it really would like i have already done like so much with this online working backwords would be different

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ordinarily 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
 3 years ago
Best ResponseYou've already chosen the best response.0or (0.71, 0.71) to the nearest hundredth.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hope it helps :) good luck!
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