Hello s_ingram Originally Posted by

**s_ingram** Hi folks,

SO, here is the problem:

A reflection in the line y = x - 1 is followed by an anticlockwise rotation of $\displaystyle 90^o $ about the point (-1,1). Show that the resultant transformation has an invariant line, and give the equation of this line. Describe the resultant transformation in relation to this line.

For the first transformation I get :

$\displaystyle \overline{x'} = \left( \begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array} \right). \overline{x}$

For the second the rotation about the origin O is:

$\displaystyle \overline{x'} = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right). \overline{x}$

and about (-1,1) is:

$\displaystyle \overline{x'} =

\left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)

\left( \begin{array}{c} x \\ y \end{array} \right) +

\left( \begin{array}{c} 0 \\ 2 \end{array} \right) $

The resultant of a process P that transforms x to x' (i.e. x' = Px) and is followed by a process Q that transforms x' to x'' (i.e. x'' = Qx') is the resultant process x'' = QPx. So:

$\displaystyle \overline{x''} =

\left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)

\left( \begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array} \right)

\left( \begin{array}{c} x \\ y \end{array} \right) +

\left( \begin{array}{c} 0 \\ 2 \end{array} \right) $

$\displaystyle \overline{x''} =

\left( \begin{array}{cc} 0 & 1 \\ 1 & 2 \end{array} \right)

\left( \begin{array}{c} x \\ y \end{array} \right) +

\left( \begin{array}{c} 0 \\ 2 \end{array} \right) $

The result of this transformation on the unit square OABC is as follows:

O(0,0) goes to O(0,2)

A(1,0) A'(0,3)

B(0,2) B'(1,4)

C(1,1) C'(1,5)

I have attached a rough sketch. It is not what I was hoping for. The answer is that the transform is a glide reflection in the line x = 1/2 with ((1/2,0) going to (1/2, 3). My transformation doesn't even preserve a closed shape. If you ignore the sequence of the letters OABC it looks like a shear along x = 1/2. Can anyone see what I am doing wrong?

Thanks for showing us your full working. I think it's pretty well correct, except for the first transformation matrix. This doesn't represent the reflection, does it? Don't forget that a transformation represented by a single 2 x 2 matrix will always leave the origin fixed; and you want $\displaystyle (0, 0) \to (1, -1)$.

You have to reflect first in a parallel line through the origin (that's $\displaystyle y = x$) and then translate so that the origin moves to the required image. So that's:$\displaystyle \overline{x'} = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array} \right). \overline{x}+\left( \begin{array}{c} 1 \\ -1 \end{array} \right) $

Do you want to try again?

Grandad