5.
Gọi G là trọng tâm tam giác PQR, theo công thức trọng tâm: \(\left\{{}\begin{matrix}x_G=\dfrac{-3+1+2}{3}=0\\y_G=\dfrac{2+1-4}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow G\left(0;-\dfrac{1}{3}\right)\)
\(V_{\left(O;-\dfrac{1}{3}\right)}\left(G\right)=G'\left(x';y'\right)\Rightarrow\left\{{}\begin{matrix}x'=-\dfrac{1}{3}.0=0\\y'=-\dfrac{1}{3}.\left(-\dfrac{1}{3}\right)=\dfrac{1}{9}\end{matrix}\right.\)
\(\Rightarrow G'\left(0;\dfrac{1}{9}\right)\)
6.
Theo công thức phép vị tự, \(V_{\left(A;k\right)}\left(B\right)=C\Rightarrow\left\{{}\begin{matrix}-1-0=k\left(2-0\right)\\5-3=k\left(-1-3\right)\end{matrix}\right.\)
\(\Rightarrow k=-\dfrac{1}{2}\)
8.
\(V_{\left(I;-2\right)}\left(d\right)=d'\Rightarrow d'\) cùng phương d
Hay pt d' có dạng: \(2x+y+c=0\) (1)
Lấy \(A\left(0;4\right)\in d\), gọi \(V_{\left(I;-2\right)}\left(A\right)=A'\left(x';y'\right)\Rightarrow A'\in d'\)
\(\left\{{}\begin{matrix}x'=-2\left(0+1\right)-1=-3\\y'=-2\left(4-2\right)+2=-2\end{matrix}\right.\)
Thế vào (1)
\(\Rightarrow2.\left(-3\right)-2+c=0\Rightarrow c=8\)
\(\Rightarrow2x+y+8=0\)
9.
\(V_{\left(O;-\dfrac{2}{3}\right)}\left(d\right)=d'\Rightarrow d'\) cùng phương d hay pt d' có dạng:
\(3x-y+c=0\) (1)
Chọn \(A\left(0;-5\right)\) là 1 điểm thuộc d
\(V_{\left(O;-\dfrac{2}{3}\right)}\left(A\right)=A'\left(x';y'\right)\Rightarrow A'\in d'\)
\(\left\{{}\begin{matrix}x'=-\dfrac{2}{3}.0=0\\y'=-\dfrac{2}{3}.\left(-5\right)=\dfrac{10}{3}\end{matrix}\right.\)
Thế vào (1):
\(3.0-\dfrac{10}{3}+c=0\Rightarrow c=\dfrac{10}{3}\)
\(\Rightarrow3x-y+\dfrac{10}{3}=0\)
\(\Leftrightarrow9x-3y+10=0\)