Rotation about a point other than the origin

Question

It is challenging for me to see the rotation of image ABCD to get to A',B',C',D'. It is easy to see the translation of the prime figure to the double prime, but not so much the pre-image to the prime figure. Is there a mathematical way to see that this is a rotation 90 CW about the point (0,2)????

Answer 1

Begin by noting that if you have a vector $\vec{v}= (a,b)$ then, a $90^{\circ}$ clock wise rotation would give the vector $\vec{v}'=(b,-a)$. A simple sketch confirms that. Also, the dot product $\vec{v} \cdot \vec{v}'=ab-ba=0$ which confirms they are perpendicular.

Now let's say the vectors for $A,B,C,D$ from $(0,2)$ are $\vec{a}, \vec{b}, \vec{c}, \vec{d}$. For the sake of an example, I'll assume (by looking at your figure) that $\vec{a}=(2,-1), \vec{b}=(4,0), \vec{c}=(5,-3), \vec{d}=(3,-5)$. Now in order to rotate these vectors $90^{\circ}$, you use the method I described above. For instance, a rotation of $\vec{a}$ about the point $(0,2)$ is $\vec{a}'=(-1,-2)$. Again remember we say that these vectors start at $(0,2)$ So using this technique, you can find $\vec{a}', \vec{b}', \vec{c}', \vec{d}'$ which are the images of $A,B,C,D$ after they've been rotated $90^{\circ}$ about $(0,2)$. You can then draw those vectors and you'll have your rotated quadrilateral.

This method can be generalized. Also consider translating the entire coordinate axis so that $(0,2)$ becomes your coordinate axis origin. Different ways to go about doing this.

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Edit:

Let's say we have a point on the quadrilateral $A$ which has a position vector $\vec{a_p}=(a_{px},a_{py})$, and let us say that this vector is rotated $90^{\circ}$ CW around some point $P$ which the position vector $\vec{p}=(p,q)$ where $p$ and $q$ is what we are after. The resultant vector after the rotation around $P$ has a position vector $\vec{a_p}'=(a_{px}',a_{py}')$, i.e. the position vector of $A'$. Also let the vector that $P$ to $A$ be $\vec{a}$ and the rotated one from $P$ to $A'$ is $\vec{a}'$ (as we did in the previous part of my answer). We also know that if $\vec{a}=(a,b)$, then $\vec{a}'=(b,-a)$.

A simple sketch should convince you that $\vec{p}+\vec{a}=\vec{a_p}$ and $\vec{p}+\vec{a}'=\vec{a_p}'$ and thus $\vec{a}=\vec{a_p}-\vec{p}=(a_{px}-p,a_{py}-q)$ and $\vec{a}'=\vec{a_p}'-\vec{p}=(a_{px}'-p,a_{py}'-q)$. Because of the relationship between $\vec{a}$ and $\vec{a}'$, we get the system: $$a_{py}-q=a_{px}'-p \\ a_{px}-p=-(a_{py}'-q)$$ which you can solve for $p$ and $q$ thus finding the coordinates of the point $P$.

Answer 2

The way I figured it out was looked at $A$s coordinate and $A'$ coordinate:

$(2, 1) \to (-1, 0)$. Knowing that a $90^{\circ}$ counterclockwise turn about the origin from $(2, 1)$ should give you $(-1, 2)$, not $(-1, 0)$, you can surmise that the point of rotation is not the origin but from the point $(0, 2)$.

Answer 3

You can use perpendicularity to find the center of rotation since the rotation is by 90 degrees. We'll start by finding the slope and midpoint of a segment joining a point on ABCD and its image (segment BB', for example). A perpendicular line through the midpoint will pass through the center of rotation. Do this twice and the intersection of the two lines will be the point you're looking for. I will provide some steps for solving algebraically, but I suspect that in most cases it will be easier to sketch the perpendicular lines on the graph itself

I will skip AA' because it's midpoint does not fall on even numbers on the grid.

BB': B(4,2) and B'(0,-2) mdpt: (2,0) slope: m = 1 perpendicular slope: m=-1

A line through (2,0) with a slope of -1 will pass through the center of rotation. If you want to use algebra rather than sketch it on the graph, the line's equation is $y=-x+2$

But which point on the line? Let's repeat with CC'...

CC': C(5,-1) and C'(-3,-3) mdpt: (1,-2) slope: m = 1/4 perpendicular slope: m = -4

A line through (1,-2) with a slope of -4 will ALSO pass through the center of rotation, so the intersection will be the point you are looking for. The line's equation is: $y=-4x+2$

Algebraically, substitute to set the expressions equal to one another:

$$-x+2=-4x+2$$ $$3x+2=2$$ $$3x=0$$ $$x=0$$

Substitute this value for x in either equation.

$$y=-(0)+2$$ $$y=2$$

Center of rotation: (0,2)