Bob sent us his solution:
Firstly I multiplied the matricies to find the new point. Using the trigonometric
identities to simplify, I got the new point as(rcos(q+f),rsin(q+f)).
This meant that we had rotated the point f degrees anticlockwise. To prove that OX = OX¢ = p, I drew a line XX', which intersects and is parallel
to the line y=xtanq (call this point of intersection D). ButDX¢=DX
and soODX and ODX¢ are two right-angled triangles of the same size, soOX=OX¢=p.
By the same argument I drew lines OP and OP¢, and so got right-angled triangles
again, so OP=OP¢=q.
By looking at the right-angled triangle OAX¢, with the angle at O being 2q,
I knew that:
cos2q =
OA¢OX
=
OA¢p
and so OA¢=pcos2q.
I then looked at the right-angled triangle X¢BP, and since the angle atX
is 2q, BP¢=qsin2q. By applying Pythagoras' Theorem to the
right-angled triangle OAX,AX¢=psin2q. Finally, applying Pythagoras'
Thereom to the triangle X¢BP¢ I found BX¢=qcos2q.
Looking at the change in X co-ordinates, I found
P¢=(pcos2q+qsin2q,psin2q-qcos2q).
So the matrix
for the reflection would be:
T =
æ ç
è
cos 2q
sin2q
sin2q
-cos2q
ö ÷
ø
You may also like to look at the
problem 'The Matrix' from
July 2003 and its solution for an explanation of how a
transformation of the plane is given by a matrix and how you
can find the image of a point by multiplying its vector by the
matrix of the transformation.