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I feel like an idiot for asking, I got this question in a practice paper, it's the first question so it's easy. enter image description here

The question is "describe fully the single transformation that will map shape $P$ onto shape $Q$"

To me this is simply "Rotation $180^o$ about the point $(0, 1)$" However the mark scheme thinks it is $90^o$ clockwise?

Is there any way this could be true, it's not the first mistake in the mark scheme, but I need to be sure for my exam. (sorry if this question is to basic for here)

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  • $\begingroup$ The acute (45 degree) angle in shape P is "pointing" to the right, and the corresponding angle in shape Q is pointing down, so it looks like the angle of rotation must be 90 degrees clockwise. Another clue: if you rotate 180 degree about the point (0,1) then that pointy corner of shape P at (4,2) would move to (-4,0), and there's nothing there. $\endgroup$ Jun 4, 2011 at 10:22

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I assume that by "transformation" they mean "isometry" (else there would be infinitely many different transformations mapping these shapes onto each other).

The mark scheme is right; this is a rotation through $90^\circ$ clockwise, but about the point $(-2,3)$, not $(0,1)$. My grasp of what may have made you think it's $180^\circ$ is probably insufficient for me to say anything helpful about that, but in case you were focussing on the points at $(1,2)$ and $(-1,0)$, which are indeed related by a rotation through $180^\circ$ about $(0,1)$, note a) that these don't correspond to each other in the shapes and b) that no other pairs of points are related by this rotation.

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  • $\begingroup$ Thanks, I feel quite stupid now :) After seeing this I can't for the life of me understand how I thought it was 180 degrees :). $\endgroup$
    – Jonathan.
    Jun 4, 2011 at 10:58
  • $\begingroup$ @Jonathan: No need to feel stupid -- live and learn :-) $\endgroup$
    – joriki
    Jun 4, 2011 at 11:20
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Your mark scheme seems to be correct. Look at the small line segment (1,4)-(2,4) in the polygon. It maps to the vertical segment (-1,0)-(-1,1).

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  • $\begingroup$ Ok I see it is 90 degrees clockwise but it's not about the point (0,1). $\endgroup$
    – Jonathan.
    Jun 4, 2011 at 10:58

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