Triangle rotate using centroid

Question

I have this centroid $O (.66,-9)$, I need to rotate $45^{\circ}$, how do I do it please?

The full coordinates are $A(2,-9)$, $B(0,-8)$, $C(0, -10)$. Thanks.

Answer 1

To rotate the coordinate $(x,y)=x+yi$ about the origin by $\frac{\pi}{4}$ radians we add or subtract $\frac{\pi}{4}$ to the argument of $x+yi=re^{i \theta}$.

Now note by Euler's formula,

$$e^{\frac{\pi}{4}i}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$$

By multiplying this complex number by the complex number $x+yi$, we add $\frac{\pi}{4}$ to the argument of $(x,y)$ because $re^{i \theta}e^{i \frac{\pi}{4}}=re^{i(\theta+\frac{\pi}{4})}$. Rotating the point $(x,y)$ $45$ degrees counterclockwise.

$$\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i \right)(x+yi)=\frac{\sqrt 2}{2}(x-y)+i\frac{\sqrt 2}{2}(x+y)$$

So the point $(x,y)$ becomes $(\frac{\sqrt 2}{2}(x-y),\frac{\sqrt 2}{2}(x+y))$ after rotation by $45$ degrees counterclockwise about the origin.

Redefine the centroid to be the origin by subtracting $.66$ from every $x$ coordinate and adding $9$ from every $y$ coordinate.

Now rotate everything about the origin (the centroid).

Work:

First we redefine the centroid to be the origin moving all points with it (these are approximations):

$$O (.66,-9) \mapsto (.66-.66,-9+9)=(0,0)$$

$$A (2,-9) \mapsto (2-.66,-9+9)=(1.33,0)$$

$$B (0,-8) \mapsto (0-.66,-8+9)=(-.66,1)$$

$$C (0,-10) \mapsto (0-.66,-10+9)=(-.66,-1)$$

Now we rotate with formula $(x,y) \mapsto (\frac{\sqrt 2}{2}(x-y),\frac{\sqrt 2}{2}(x+y))$,

$$O \mapsto (0,0)$$

$$A \mapsto (\frac{\sqrt{2}}{2}(1.33), \frac{\sqrt{2}}{2}(1.33))$$

$$B \mapsto (\frac{\sqrt{2}}{2}(-1.66), \frac{\sqrt{2}}{2}(.33))$$

$$C \mapsto (\frac{\sqrt{2}}{2}(.33),\frac{\sqrt{2}}{2}(-1.66))$$

Now we undo the translation we did in the first plane by adding $.66$ and subtracting $9$ to the $x$ coordinate and $y$ coordinate respectively.

To get,

$$A'=(\frac{\sqrt{2}}{2}(1.33)+.66, \frac{\sqrt{2}}{2}(1.33)-9)$$

$$B'=(\frac{\sqrt{2}}{2}(-1.66)+.66, \frac{\sqrt{2}}{2}(.33)-9)$$

$$C'=(\frac{\sqrt{2}}{2}(.33)+.66,\frac{\sqrt{2}}{2}(-1.66)-9)$$