Lấy các điểm $A(1;0)$ và $B(-2;2)\in d$
$A'\Big(1\cos60^o; 1\sin60^o\Big)=\Big(\dfrac{1}{2};\dfrac{\sqrt3}{2}\Big)$
$B'\Big(-2\cos60^o-2\sin60^o; -2\sin60^o+2\cos60^o\Big)=\Big(-1-\sqrt3;1-\sqrt3\Big)$
$\Rightarrow \vec{A'B'}\Big(\dfrac{-3-2\sqrt3}{2};\dfrac{2-3\sqrt3}{2})$
$\Rightarrow \vec{u}=k\vec{A'B'}=(-3-2\sqrt3; 2-3\sqrt3)$
$d': (-3-2\sqrt3)(x-\dfrac{1}{2})+(2-3\sqrt3)(y-\dfrac{\sqrt3}{2})=0$
$\Leftrightarrow (-3-2\sqrt3)x+(2-3\sqrt3)y+6=0$
