Actually, I figured it out on my own. But now I have a different problem. I am plotting an ellipse in Matlab, and I want to rotate it 45 degrees, now clockwise, around it's center. Any ideas?
I don't know if this is the right forum, but here we go.
I have the situation where I have several known points in the plane that I want to rotate 45 degrees counterclockwise around a known center which is not (0,0). How do I do this, i.e. how do I find the new coordinates for each point?
I am using Matlab, if that matters.
Say the point is (1,1).
And we want to find how it rotates ("a rotation matrix", but if that terms confuses you, ignore it).
Draw a new coordinate system having (1,1) in its center pararrel and perpindicular to the original system.
Then point (x,y) relative to the new coordinate system gets mapped to,
x'=sqrt{2}/2*x-sqrt{2}/2*y
y'=sqrt{2}/2*x+sqrt{2}/2*y
Thus, (x',y') is the new coordinate relative to the new coordinate system.
But relative to the old coordinate system it is,
(x'+1,y'+1)
Hello, changing_seasons!
I think I've worked this out correctly . . .
Let P(x,y) be any point.I have several known points in the plane that I want to rotate 45° CCW
around a known center which is not (0,0).
How do I find the new coordinates for each point?
Let C(h,k) be the given center.
We will need two quantities:
. . . . . . . . . . . . - - - . . .______________
The distance CP: . r .= .√(x - h)² + (y - k)²
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . y - k
The angle θ that CP makes with the positive x-axis: . tan θ .= .------
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x - h
Then: . x' .= .h + r·cos(θ + 45°)
. . . . . .y' .= .k + r·sin(θ + 45°)
. . I think . . .