trong mp oxy , cho d:x-2y+3=0. pt ảnh của d qua phép quay tâm o góc quay pi/3 là 29/08/2021 Bởi Arya trong mp oxy , cho d:x-2y+3=0. pt ảnh của d qua phép quay tâm o góc quay pi/3 là
Lấy hai điểm $A(-3;0)$, $B(1;2)\in d$ $Q_{(O;\dfrac{\pi}{3})}: A\to A’, B\to B’; d\to d’$ Ta có: $x_{A’}= -3\cos\dfrac{\pi}{3}-0.\sin\dfrac{\pi}{3}=-\dfrac{3}{2}$ $y_{A’}=-3\sin\dfrac{\pi}{3}+0.\cos\dfrac{\pi}{3}=\dfrac{-3\sqrt3}{2}$ $\Rightarrow A'(\dfrac{-3}{2};\dfrac{-3\sqrt3}{2})$ $x_{B’}= 1.\cos\dfrac{\pi}{3}-2\sin\dfrac{\pi}{3}= \dfrac{1-2\sqrt3}{2}$ $y_{B’}= 1.\sin\dfrac{\pi}{3}+2\cos\dfrac{\pi}{3}=\dfrac{2+\sqrt3}{2}$ $\Rightarrow B'(\dfrac{1-2\sqrt3}{2};\dfrac{2+\sqrt3}{2})$ $\Rightarrow \vec{A’B’}=(2-\sqrt3; 1+2\sqrt3)=\vec{u_{d’}}$ $\Rightarrow \vec{n_{d’}}=(1+2\sqrt3; \sqrt3-2)$ $(d’): (1+2\sqrt3)(x+\dfrac{3}{2})+(\sqrt3-2)(y+\dfrac{3\sqrt3}{2})=0$ $\Leftrightarrow (1+2\sqrt3)x+(\sqrt3-2)y+6=0$ Bình luận
Lấy hai điểm $A(-3;0)$, $B(1;2)\in d$
$Q_{(O;\dfrac{\pi}{3})}: A\to A’, B\to B’; d\to d’$
Ta có:
$x_{A’}= -3\cos\dfrac{\pi}{3}-0.\sin\dfrac{\pi}{3}=-\dfrac{3}{2}$
$y_{A’}=-3\sin\dfrac{\pi}{3}+0.\cos\dfrac{\pi}{3}=\dfrac{-3\sqrt3}{2}$
$\Rightarrow A'(\dfrac{-3}{2};\dfrac{-3\sqrt3}{2})$
$x_{B’}= 1.\cos\dfrac{\pi}{3}-2\sin\dfrac{\pi}{3}= \dfrac{1-2\sqrt3}{2}$
$y_{B’}= 1.\sin\dfrac{\pi}{3}+2\cos\dfrac{\pi}{3}=\dfrac{2+\sqrt3}{2}$
$\Rightarrow B'(\dfrac{1-2\sqrt3}{2};\dfrac{2+\sqrt3}{2})$
$\Rightarrow \vec{A’B’}=(2-\sqrt3; 1+2\sqrt3)=\vec{u_{d’}}$
$\Rightarrow \vec{n_{d’}}=(1+2\sqrt3; \sqrt3-2)$
$(d’): (1+2\sqrt3)(x+\dfrac{3}{2})+(\sqrt3-2)(y+\dfrac{3\sqrt3}{2})=0$
$\Leftrightarrow (1+2\sqrt3)x+(\sqrt3-2)y+6=0$