You are watching: How to rotate a figure around a point not the origin
Begin by noting the if you have actually a vector $\vecv= (a,b)$ then, a $90^\circ$ clock wise rotation would provide the vector $\vecv"=(b,-a)$. A simple sketch confirms that. Also, the period product $\vecv \cdot \vecv"=ab-ba=0$ which confirms they are perpendicular.
Now let"s to speak the vectors because that $A,B,C,D$ from $(0,2)$ room $\veca, \vecb, \vecc, \vecd$. Because that the services of an example, I"ll assume (by spring at your figure) that $\veca=(2,-1), \vecb=(4,0), \vecc=(5,-3), \vecd=(3,-5)$. Currently in order to revolve these vectors $90^\circ$, you use the technique I described above. Because that instance, a rotation that $\veca$ about the allude $(0,2)$ is $\veca"=(-1,-2)$. Again remember we say the these vectors begin at $(0,2)$ So making use of this technique, friend can uncover $\veca", \vecb", \vecc", \vecd"$ which space the photos of $A,B,C,D$ after they"ve been rotated $90^\circ$ about $(0,2)$. You can then draw those vectors and also you"ll have actually your rotated quadrilateral.
This an approach can it is in generalized. Also consider translating the entire coordinate axis so the $(0,2)$ becomes your coordinate axis origin. Various ways come go around doing this.
Let"s to speak we have a allude on the square $A$ which has actually a position vector $\veca_p=(a_px,a_py)$, and also let united state say that this vector is rotated $90^\circ$ CW around some suggest $P$ i m sorry the position vector $\vecp=(p,q)$ whereby $p$ and also $q$ is what we are after. The result vector after the rotation roughly $P$ has a place vector $\veca_p"=(a_px",a_py")$, i.e. The position vector of $A"$. Likewise let the vector that $P$ to $A$ be $\veca$ and also the rotated one native $P$ come $A"$ is $\veca"$ (as us did in the previous part of my answer).
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We also know the if $\veca=(a,b)$, climate $\veca"=(b,-a)$.
A simple sketch need to convince you that $\vecp+\veca=\veca_p$ and also $\vecp+\veca"=\veca_p"$ and also thus $\veca=\veca_p-\vecp=(a_px-p,a_py-q)$ and also $\veca"=\veca_p"-\vecp=(a_px"-p,a_py"-q)$. Due to the fact that of the relationship between $\veca$ and $\veca"$, we gain the system: $$a_py-q=a_px"-p \\ a_px-p=-(a_py"-q)$$ i m sorry you deserve to solve for $p$ and $q$ therefore finding the coordinates of the point $P$.