Trong mặt phẳng $oxy$,cho $2$ đường thẳng $(d):3x+2y-1=0 $và $(d'):6x+4y+3=0,$ và điểm $A(1;2) $

a,viết phương trình đường thẳng $\triangle $ là ảnh của đường thẳng $(d) $qua phép $V_{(O,-2)}$

b, xác định phép vị tự $V_{(A)}$ biến đường thẳng $(d) $thành đường thẳng $(d')$?
bạn gõ đề lại nhé, ko hiểu –  HọcTạiNhà 07-12-13 04:34 AM
Lấy $M(1;\ -1) \in (d) $ Gọi $M'(x;\ y) \in (d')$ với $V_{(A;\ k)} (M) \longrightarrow M'$

Mà $V_{(A;\ k)} (d) \longrightarrow (d') \Rightarrow M' \in (d')$

Ta có $\vec{AM'}=k\vec{AM} \Leftrightarrow \begin{cases} x-1 =k(1-1) \\ y-2 =k(-1-2) \end{cases} \Leftrightarrow M'(1;\ 2-3k)$

$M' \in (d') \Leftrightarrow 6.1 + 4.(2-3k)+3=0 \Leftrightarrow k=\dfrac{17}{12}$

Vậy phép vị tự tâm $A$ tỉ số $k=\dfrac{17}{12}$ biến $(d)$ thành $(d')$
a/ Ta có: $V_{(O;-2)}(d)=\Delta \Rightarrow \Delta :3x+2y+m=0$
- Lấy $M(1;-1)\in d$
$- V_{(O;-2)} (M)=M'(x;y)\in \Delta \leftrightarrow \overrightarrow{OM'}=-2\overrightarrow{OM} \Leftrightarrow \begin{cases}x= -2\\ y= 2\end{cases}\Rightarrow M'(-2;2)$
- vì $M'\in \Delta $ nên ta có: $3.(-2)+2.2+m=0\Leftrightarrow m=2$
Vậy:  $\Delta :3x+2y+2=0.$

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