Question

With a two dimensional surface, if we take (2,1) as the center point and consider a transformation witha rotation angle of 45º, then point (3,3) is transformed into point ??

Answer 1

Sketching this situation could be very helpful. Plot (2,1). Draw a horiz. line thru (2,1). Rotate this line counterclockwise by 45 degrees.

To be honest, I don't yet know where that point (3,3) goes.

Try this: plot both (2,1) and (3,3). Connect these two points with a line segment.

Now rotate the whole works about the origin. I'm frankly guessing here regarding what to do. Perhaps you could come up with alternative ideas.

Answer 2

Applying complex numbers can be useful for this kind of thing.

The complex number z = x + yi representing the point (x,y) when rotated by 45 degrees counter clockwise is the same as
\[z_T = z (\cos \frac{ \pi }{ 4 } + i \sin \frac{ \pi }{ 4 }) = z(\frac{ \sqrt{2} }{ 2} + i \frac{ \sqrt{2} }{ 2 })\]

For (3,3) we have z = 3+3i, its modulus is \[|z| = \sqrt{3^2+3^2} = 3 \sqrt{2}\] and its argument is \[Arg(z) = \frac{ \pi }{ 4 }\]
Thus its modulus argument form is \[3\sqrt{2} ~cis ~\frac{ \pi }{ 4 }\]
Multiplying this by cis(pi/4) gives you 3 sqrt(2) cis pi/2 = 3sqrt(2) i which is just y = 3sqrt(2).

now if you don't know complex numbers, then this is perhaps the most straight forward approach:

Draw P(3,3) on Cartesian plane. The gradient of OP is 1 so the angle it makes with the horizontal is 45 degrees. If you turn it counterclockwise 45 degrees then it becomes 90 degrees perpendicular to the x axis. i.e. the new point lies on the y axis.
Now we need its length.

Length is the same as the length of OP. By Pythagoras' theorem

this is
\[\sqrt{3^2+3^2} = 3 \sqrt{2}\]

So (3,3) becomes (0, 3sqrt(2)) on our new reference system.