\(x'=x\cos\theta-y\sin\theta\\y'=x\sin\theta+y\cos\theta\\\\
x'=2\cdot\cos 240-3\cdot\sin240\\
x'=2\cdot(-\frac{1}{2})-3\cdot(-\frac{\sqrt3}{2})\\
x'=-1+\frac{3\sqrt3}{2}\\
y'=2\cdot\sin 240+3\cdot\cos 240\\
y'=2\cdot(-\frac{\sqrt3}{2})+3\cdot(-\frac{1}{2})\\
y'=-\sqrt3-\frac{3}{2}\\\\
(-1+\frac{3\sqrt3}{2},-\sqrt3-\frac{3}{2})\)
