1. ## HELP: Rotating Figures

Hi

These four points A(3,1), B(5/2,2), C(9/2,3), D(5,2)

Is turned 90 degrees clockwise around the center of the figure.

I'm supposed to write a transformation that does this.

I get the following:

$(x,y) \rightarrow \left[ \begin{array}{cc} -cos( 2 \theta ) & sin(2 \theta) \\ sin(2 \theta) & cos(2 \theta) \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right]$

My teacher is says this is wrong cause any transformations that isn't rotated around (0,0) is not a linear transformation.

If thats true what should the transformation then look like ???

/Fred

2. Originally Posted by Mathman24
[These four points are to be rotated] 90 degrees clockwise around the center of the figure.
Let's say a rotation of the desired form abou the origin is of the form $\bold{y} = A \bold{x}$ where A is a matrix and x and y are 2-vectors. You'll have to determine the matrix A - your solution isn't correct since it doesn't pay attention to the 90 degrees. Such a rotation is linear.

Suppose now you want to rotate about a different point, say $\bold{x_0}$. Set $\bold{x'} = \bold{x - x_0}$. This is the displacement of the point x from $\bold{x_0}$.

Then $\bold{y'} = A\bold{x'} = A(\bold{x-x_0})$ is the rotated position vector, relative to the point $\bold{x_0}$, and therefore $\bold{y} = \bold{y'+x_0} = A\bold{(x-x_0)}+\bold{x_0}$ is the formula for the rotation about $\bold{x_0}$.

3. Hi can't it be made correct if I add the distance between the original point and the center of the figure ??

/Fred

Originally Posted by hpe
Let's say a rotation of the desired form abou the origin is of the form $\bold{y} = A \bold{x}$ where A is a matrix and x and y are 2-vectors. You'll have to determine the matrix A - your solution isn't correct since it doesn't pay attention to the 90 degrees. Such a rotation is linear.

Suppose now you want to rotate about a different point, say $\bold{x_0}$. Set $\bold{x'} = \bold{x - x_0}$. This is the displacement of the point x from $\bold{x_0}$.

Then $\bold{y'} = A\bold{x'} = A(\bold{x-x_0})$ is the rotated position vector, relative to the point $\bold{x_0}$, and therefore $\bold{y} = \bold{y'+x_0} = A\bold{(x-x_0)}+\bold{x_0}$ is the formula for the rotation about $\bold{x_0}$.