Please help me. I don't know how to do this without drawing and tracing. That is very slow. This is middle school math. Please don't use sin and cos. Please expain this for me. :confused: Thank you.
Draw triangle LMN with vertices L(2, -1), M(6, -2) and N (4, 2). Find the coodinates of the vertices after a counterclockwise 90 degree rotation about the origin and about each of the point L, M and N.
Luckily for both us, there are 90° rotations; there's a simple routine.
Suppose we have a point relative to a point
O * → → → → → *
Think of the slope of
. . That is, going from to , we move: .
Let be the new point after a 90° CCW rotation.
The slope of is perpendicular to
. . Hence: .
That is, going from to , we move: .
We know that will be northwest of
. . so we use the second fraction: .
From point , we move: .left 2, up 3.
* ← ← ← * (3,2)
O * - - - - - *
And we arrive at: .
Do a few more of these and the steps will get smoother.
You'll come up with your own intuitive shortcuts.
Trust me on this . . .
I can love math again!
Thank you!Thank you! Thank you! You are sooooo clear, I got it!(Clapping)
Am I right?
Please clear this up for me. :confused:
P rotates around O 90 degrees CCW, if I let O to be the origin, after the rotation P' will be (-2, 3) (I tried this and used tracing method.).
In your calculation, you didn't write the coordiantes of O. Does this mean O can be any point? But how come it does not work for (0, 0).
I looked at the tracing, the slopes of OP and OP' seem to perpendicular.
What did I do wrong?(Worried)