Respuesta :

konrad509
[tex]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}) [/tex]

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