# Math Help - Transformation geometry

1. ## Transformation geometry

2. Hello walreinlim88
Originally Posted by walreinlim88
I think there is a mistake in the question. I believe the transformation matrix should be:
$\begin{pmatrix}\dfrac{\sqrt5}{3}&\dfrac23\\ \dfrac23&-\dfrac{\sqrt5}{3}\end{pmatrix}$
which represents a reflection in the line through the origin that makes an anticlockwise angle $\tfrac12\theta$ with the $x$-axis, where $\cos\theta = \frac{\sqrt5}{3}$.

See the diagram attached, where PQ is the mirror-line. Note that $A(1,0) \to A'(\sqrt5/3,2/3)$ and $B(0,1) \to B'(2/3,-\sqrt5/3)$.

It's fairly straighforward to show that if $\cos\theta = \frac{\sqrt5}{3}$, then $\tan\tfrac12\theta = \frac{3-\sqrt5}{2}$, and hence the equation of the mirror (invariant) line is
$y=\frac{3-\sqrt5}{2}x$
(Use $\cos^2\tfrac12\theta = \tfrac12(1+\cos\theta)$ and $\tan\tfrac12\theta = \sqrt{\sec^2\tfrac12\theta -1}$.)

Can you continue from here?

3. how about 2nd part? i dont know how to find the point of T.

4. Hello walreinlim88
Originally Posted by walreinlim88
how about 2nd part? i dont know how to find the point of T.
Find the image of one of the points in the mirror-line; join its reflection to the other point. T is the point where this line intersects the mirror-line.