I have a rectangle in the cartesian plane defined by the top left and Bottom right as $(3, 5),(5, 3)$
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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$\begingroup$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 $(5,-5)$.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ $\begingroup$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 $(3,-3),(5,-3),(5,-5),(3,-5)$
$\endgroup$ 2 $\begingroup$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.
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