A đối xứng vs B qua K=> \(K\left(x_K;y_K\right)\in\left(d\right):y=\frac{x+3}{2}=\frac{1}{2}x+\frac{3}{2}\)
\(\overrightarrow{AK}=\overrightarrow{KB}\Leftrightarrow\left(x_K-x_A;y_K-y_A\right)=\left(x_B-x_K;y_B-y_K\right)\)
\(\Leftrightarrow\left(x_K-x_A;y_K\right)=\left(-x_K;y_B-y_K\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x_K=x_A\\2y_K=y_B\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x_K=x_A\\2.\frac{1}{2}x_K+2.\frac{3}{2}=y_B\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_K=\frac{x_A}{2}\left(1\right)\\\frac{x_A}{2}+3=y_B\left(2\right)\end{matrix}\right.\)
\(AB\perp\left(d\right)\)
\(\Leftrightarrow\overrightarrow{AB}.\overrightarrow{OK}=0\)
\(\Leftrightarrow\left(x_B-x_A;y_B-y_A\right)\left(x_K;y_K\right)=0\)
\(\Leftrightarrow-x_A.x_K+y_B.y_K=0\)
\(\Leftrightarrow y_B.\frac{1}{2}x_K+y_B.\frac{3}{2}-x_A.x_K=0\left(3\right)\)
Thay (1) vào (3):
\(\Rightarrow y_B.\frac{1}{2}.\frac{x_A}{2}+y_B.\frac{3}{2}-x_A.\frac{x_A}{2}=0\)
\(\Leftrightarrow\frac{x_A.y_B}{4}+\frac{3}{2}y_B-\frac{x_A^2}{2}=0\left(4\right)\)
Rồi ok đến đây cậu tự giải nốt bằng cách giải hpt (2) và (4) là ra