Rotate triangle ABC around the origin

Question

How do you get from triangle ABC (blue) to triangle ABC (red)

The instructions are to rotate it $270$ degrees

I am trying to help a friend but forgot how to do this? Is it using a formula?

Answer 1

Think of the vertices as being complex numbers.

1. $A=-2+3\,i$
2. $B=-2+0\,i$
3. $C=-4+2\,i$

Each point may be rotated $270^\circ$ or $\frac{3\pi}{2}$ radians about the pole $0$ by multiplying by

$$e^{\frac{3\pi}{2}i}=\cos\left(\frac{3\pi}{2}\right)+i\,\sin\left(\frac{3\pi}{2}\right)=-i$$

This gives

1. $A^\prime=3+2\,i$
2. $B^\prime=2\,i$
3. $C^\prime=2+4\,i$

Answer 2

$\triangle ABC$ $ \rightarrow$ $\triangle A'B'C'.$

$x,y$ $\rightarrow$ $ x',y'$, a rotation about the origin by $\alpha =90°$, clockwise:

1)$x' = x\cos(\alpha) +y\sin(\alpha)$.

2)$y' = -x\sin(\alpha) + y\cos(\alpha)$,

$\alpha = 90°.$

$A(-2,3)$ $ \rightarrow$ $A'(3,2)$.

$B(-2,0)$ $\rightarrow$ $B'(0,2).$

$C(-4,2)$ $\rightarrow$ $C'(2,4).$

Can you derive 1),2) from scratch?