Timtime 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, ?

1. JakeV8

You don't even have to rotate the triangle... just rotate the point P.

2. AbhimanyuPudi

y coordinate is also the same..

3. Timtime

so you would basically switch the x points to y points?

4. JakeV8

|dw:1349722508665:dw|

5. JakeV8

no, you can't swap coordinates if you only rotate 45 degrees.

6. JakeV8

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

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

You are doing good dude @JakeV8 I'll leave it to you..

9. JakeV8

@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

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

|dw:1349722859375:dw|

12. Timtime

were taught to use use matrices to solve the problem {0 -1} {1 0}

13. JakeV8

interesting. Is a 45 degree rotation something you can express as matrix transformation?

14. Timtime

yes its a matrix rotation

15. JakeV8

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

its weird how my online class is doing it

17. JakeV8

Does knowing the answer help you figure out how to set up the matrix rotation, working backward?

18. Timtime

it really would like i have already done like so much with this online working backwords would be different

19. JakeV8

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

or (-0.71, 0.71) to the nearest hundredth.

21. Timtime

thanks you

22. JakeV8

hope it helps :) good luck!

23. afranklin12

ur cute