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I have a rectangle in the cartesian plane defined by the top left and Bottom right as $(3, 5),(5, 3)$

the picture

We rotate this around the origin clockwise by 90 degrees, what is the new top left and bottom right point?

Is it: $(3, -3), (5, -5)$ ?

This doesnt graphically make sense to me though?

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    What specifically doesn't make sense? – 2017-01-26
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    How can the top left y point become negative? @TheCount. Is it correct tho? – 2017-01-26
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    How much are you rotating it around the origin? – 2017-01-26
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    I have no idea, because I don't know how much you rotate by. – 2017-01-26
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    @HarnoorLal read the title. – 2017-01-26
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    Rotating 90 degrees will bring the opposite corners to top left and bottom right. In other words, the image of the points you gave for the original position will not be at top left and bottom right any more. However the negative y-coordinates are no mystery - the rotation has taken the figure from Quadrant I to Quadrant IV – 2017-01-26
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    @Joffan, so is my answer accurate? I did switch the coordinates – 2017-01-26
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    Yes, the answer is correct - I thought you knew that, and were asking for explanation and insight. – 2017-01-26
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    @Joffan, someone answered saying my answer is wrong? – 2017-01-26
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    There is some ambiguity about what is required, which will lead to some different answers, but I think you are right on my interpretation of the question. – 2017-01-26
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    NOTE: The y-coordinate of EVERY SINGLE POINT in the rectangle becomes negative after this rotation, because the whole rectangle is moved into the 4th quadrant (to the right of the y-axis and below the x-axis). – 2017-01-26
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    @BrevanEllefsen That is an edited title. It didn't specify before. – 2017-01-27

4 Answers 4

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Your rectangle is actually a square. The other two corners are at $(3,3)$ and $(5,5)$, on the $45^{\circ}$ diagonal (from the origin to the upper right). Rotating it clockwise by $90^{\circ}$ produces a square with corners $(3,-3)$ and $(5,-5)$ on the $-45^{\circ}$ diagonal (from the origin to the lower right), so the other corners (the rotated versions of the corners you mention) are at $(5,-3)$ and $(3,-5)$, in that order.

Why are you wondering about negative coordinates? You can rotate the square to any quadrant, so you can have any combination of negative/positive coordinates. Moving clockwise, these are:

  • Quadrant I = (+,+)
  • Quadrant IV = (+,-)
  • Quadrant III = (-,-)
  • Quadrant II = (-,+)

Your rectangle moved from Quadrant I to Quadrant IV.

Addendum: After the rotation, the "new" top left is $(3,-3)$ and the "new" bottom right is $(5,-5)$.

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    I want the current top left and the current bottom right. So after the transformation it is a new rectangle, what are its top left and bottom right coordinates? Joffan said my answer was right but now there is a contradiction – 2017-01-26
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    Okay, I added that in my answer. – 2017-01-26
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Rotations in the plane by an angle $\theta$ can be calculated using the rotation matrix $$\begin{bmatrix}\cos \theta& -\sin \theta \\ \sin \theta& \cos\theta\end{bmatrix}.$$

So in the case of a rotation of $\theta=90^\circ$ this matrix is $$\begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix}.$$

So the new corners of your rectangle are $$ \begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix} \begin{bmatrix}3 \\ 5\end{bmatrix} = \begin{bmatrix}-5\\ 3\end{bmatrix}, \qquad\begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix} \begin{bmatrix}5 \\ 3\end{bmatrix} = \begin{bmatrix}-3\\ 5\end{bmatrix}, $$

$$ \begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix} \begin{bmatrix}3 \\ 3\end{bmatrix} = \begin{bmatrix}-3\\ 3\end{bmatrix}, \qquad\begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix} \begin{bmatrix}5 \\ 5\end{bmatrix} = \begin{bmatrix}-5\\ 5\end{bmatrix}. $$

So the cooardinate of the top left corner is $(-5,5)$ and the coordinate of the bottom right corner is $(-3,3)$

Keep in mind: Due to the rotation of the rectangle itself, the new top left corner is not the image of the old top left coordinate.

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No, after rotating your square $90^\circ$ clockwise your top left is $(5,-3)$ and your bottom right $(3,-5)$

Note: the corners refer to the ones you describe at the beginning as "top left" and "bottom right" and where do they end up if you track them. The 4 corners of the rotated square are $(3,-3),(5,-3),(5,-5),(3,-5)$

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    Those would be top right and bottom left of the rotated square. – 2017-01-26
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    well, if you stick with the original 2 corners no. If you look at the cartesian plane yes. I am referring to the corners she describes at the beginning and where do they end up if you track them. – 2017-01-26
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One thing that bugs me about this question is that the OP is using the verb "to rotate" when "to orbit" is what is meant. Rotation occurs when the object about which the parts of the figure are orbiting is also a part of the figure.

The earth cannot be said to rotate about the sun; it rotates about its polar axis and it orbits the sun. The solar system cannot be said to rotate; all of its parts move at different speeds, so the concept of rotation doesn't apply.

My next paragraph is based upon my conception of "to rotate."

The OP did say "rectangle," so the other two omitted vertices are on the circle $(x-4)^2+(y-4)^2=2.\quad$ $\theta\;$ is an arbitrary parameter. The first is $(4+\sqrt2·\cos{\theta}\mid 4+\sqrt2·\sin{\theta}).\quad$ The second is $(4+\sqrt2·\cos(\theta+\pi)\mid 4+\sqrt2·\sin(\theta+\pi)).\qquad$ This doesn't eliminate the square, of course.