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"With a 2 dimensional surface: We take (2, 1) as the center point and consider a transformation with a rotation angle of 45◦ so point (3, 3) is transformed into point ...?"

I'm really close to getting the answer! I've gotten (-1/sqrt2,3/sqrt2) but the answer is (2-1/sqrt2, 3+1/sqrt2). Please tell me what I'm missing.

Thank you!

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  • $\begingroup$ See here for advice on what is expected from askers. Also, there are some unclear (but mostly harmless) points in your question. 1) Surface? What surface? The surface of a sphere? Are you talking about the plane? Looks like you are! Possibly something lost in translation. 2) 45 degrees? Clockwise or counterclockwise? In some sources clockwise rotation would get a negative sign, but that isn't universal. $\endgroup$ – Jyrki Lahtonen Feb 17 '19 at 4:00
  • $\begingroup$ Anyway, here's a possible way of doing this if you are unfamiliar with the formula for a general rotation by angle $\theta$. It may be easier to figure out the direction of the line you get by rotating the line segment from $(2,1)$ to $(3,3)$ by 90 degrees (to whichever direction you are supposed to be rotating). And then you can bisect that angle. The points on the bisector are equidistant from the two lines, so you can derive its equation using the formula for the distance of a point from a line. Warning: there will be two bisectors. $\endgroup$ – Jyrki Lahtonen Feb 17 '19 at 4:06
  • $\begingroup$ @jyrkiLahtonen regarding to the first answer, the question says that literally. I guess that, as it says only 45 (and not -/negative) it should be a CCW rotation. $\endgroup$ – JennYT Feb 18 '19 at 14:05
  • $\begingroup$ Agreed. ${}{}{}{}$ $\endgroup$ – Jyrki Lahtonen Feb 18 '19 at 14:15
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Draw out the graph. Also draw out the situation when it is rotated $90°$, calculate the gradient of line connecting point (0,2) and (3,3), and then derive the gradient of its perpendicular line, which pass through the desired point. Finally use the gradient with the length between the centre and the desired point to obtain the coordinate of the desired point. The approach is quite tedious though..

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  • $\begingroup$ The triangle is isosceles, but the basis is not necessarily parallel to the $x$-axis. $\endgroup$ – AlessioDV Feb 17 '19 at 3:54
  • $\begingroup$ Sorry I got it wrong. Edited the answer. $\endgroup$ – qsmy Feb 17 '19 at 4:08
  • $\begingroup$ @qsmy May I know in which Math topic this is? I'd like to read about it since I don't understand much $\endgroup$ – JennYT Feb 18 '19 at 19:59
  • $\begingroup$ @JennYT It is mostly about linear graph and geometry/trigonometry. This question can be solved with this approach mainly because of the special coordinates and angle, therefore able to apply the concept of "the multiplication of the two gradients of two lines perpendicular to each other equals to 1" $\endgroup$ – qsmy Feb 19 '19 at 5:45
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Hint (assuming that you are working on $\mathbb{R}^2$). Write the $2\times 2$ matrix $R$ of the rotation of $45°$ about the origin. Let $\mathbf{c}=(2,1)\in \mathbb{R}^2$ the coordinates of the centre of the rotation. The transformation you are looking for is $$ \mathbf{x} \mapsto R(\mathbf{x}-\mathbf{c}) + \mathbf{c}. $$

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