a.
\(V_{\left(A;-\dfrac{1}{3}\right)}=B'\left(x';y'\right)\Rightarrow\left\{{}\begin{matrix}x'=5+\left(-\dfrac{1}{3}\right)\left(3-5\right)=\dfrac{17}{3}\\y'=-2+\left(-\dfrac{1}{3}\right)\left(4-\left(-2\right)\right)=-4\end{matrix}\right.\)
\(\Rightarrow B'\left(\dfrac{17}{3};-4\right)\)
b.
\(V_{\left(A;2\right)}\left(H\left(x;y\right)\right)=K\Rightarrow\left\{{}\begin{matrix}4-5=2\left(x-5\right)\\-5-\left(-2\right)=2\left(y+2\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{2}\\y=-\dfrac{7}{2}\end{matrix}\right.\)
\(\Rightarrow H\left(\dfrac{9}{2};-\dfrac{7}{2}\right)\)
c.
Gọi \(V_{\left(A;-3\right)}\left(d\right)=d'\Rightarrow d'||d\) \(\Rightarrow\) phương trình d' có dạng: \(2x+3y+c=0\) (1)
Chọn \(M\left(0;1\right)\in d\) , gọi \(V_{\left(A;-3\right)}\left(M\right)=M'\left(x';y'\right)\Rightarrow M'\in d'\)
\(\left\{{}\begin{matrix}x'=5-3\left(0-5\right)=20\\y'=-2-3\left(1+2\right)=-11\end{matrix}\right.\)
Thế vào (1): \(2.20+3.\left(-11\right)+c=0\Rightarrow c=-7\)
Phương trình d' là: \(2x+3y-7=0\)
d.
(C) có tâm \(I\left(-1;0\right)\) bán kính \(R=4\)
\(V_{\left(A;-3\right)}\left(C\right)=\left(C'\left(I';R'\right)\right)\Rightarrow\left\{{}\begin{matrix}I'\left(x';y'\right)=V_{\left(A;-3\right)}\left(I\right)\\R'=\left|-3\right|.R=12\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x'=5-3\left(-1-5\right)=23\\y'=-2-3\left(0+2\right)=-8\end{matrix}\right.\)
Phương trình (C') có dạng:
\(\left(x-23\right)^2+\left(y+8\right)^2=144\)
