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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
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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JakeV8Best ResponseYou've already chosen the best response.2
You don't even have to rotate the triangle... just rotate the point P.
 one year ago

AbhimanyuPudiBest ResponseYou've already chosen the best response.1
y coordinate is also the same..
 one year ago

TimtimeBest ResponseYou've already chosen the best response.1
so you would basically switch the x points to y points?
 one year ago

JakeV8Best ResponseYou've already chosen the best response.2
no, you can't swap coordinates if you only rotate 45 degrees.
 one year ago

JakeV8Best ResponseYou've already chosen the best response.2
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

JakeV8Best ResponseYou've already chosen the best response.2
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

AbhimanyuPudiBest ResponseYou've already chosen the best response.1
You are doing good dude @JakeV8 I'll leave it to you..
 one year ago

JakeV8Best ResponseYou've already chosen the best response.2
@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

JakeV8Best ResponseYou've already chosen the best response.2
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

TimtimeBest ResponseYou've already chosen the best response.1
were taught to use use matrices to solve the problem {0 1} {1 0}
 one year ago

JakeV8Best ResponseYou've already chosen the best response.2
interesting. Is a 45 degree rotation something you can express as matrix transformation?
 one year ago

TimtimeBest ResponseYou've already chosen the best response.1
yes its a matrix rotation
 one year ago

JakeV8Best ResponseYou've already chosen the best response.2
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

TimtimeBest ResponseYou've already chosen the best response.1
its weird how my online class is doing it
 one year ago

JakeV8Best ResponseYou've already chosen the best response.2
Does knowing the answer help you figure out how to set up the matrix rotation, working backward?
 one year ago

TimtimeBest ResponseYou've already chosen the best response.1
it really would like i have already done like so much with this online working backwords would be different
 one year ago

JakeV8Best ResponseYou've already chosen the best response.2
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

JakeV8Best ResponseYou've already chosen the best response.2
or (0.71, 0.71) to the nearest hundredth.
 one year ago

JakeV8Best ResponseYou've already chosen the best response.2
hope it helps :) good luck!
 one year ago
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