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.

**Edit:**

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 $P$.