a/ \(\overrightarrow{BC}=\left(6;-3\right)\Rightarrow\overrightarrow{n_{BC}}=\left(3;6\right)\)
\(\Rightarrow BC:3\left(x-1\right)+6\left(y+2\right)=0\)
\(BC:3x+6y+9=0\)
b/ Phương trình đường cao AA' nhận \(\overrightarrow{BC}\) làm vecto pháp tuyến\(\Rightarrow\overrightarrow{n_{AA'}}=\left(6;-3\right)\)
\(\Rightarrow AA':6\left(x-2\right)-3y=0\)
\(:6x-3y-12=0\)
c/ \(AA'\cap BC=\left\{A'\right\}\Rightarrow\left\{{}\begin{matrix}6x-3y-12=0\\3x+6y+9=0\end{matrix}\right.\Rightarrow A'\left(1;-2\right)\)
d/ \(\overrightarrow{AM}=\overrightarrow{MC}\Leftrightarrow\left(x_M-2;y_M\right)=\left(1-x_M;-2-y_M\right)\)
\(\Rightarrow M\left(\frac{3}{2};-1\right)\)
\(\overrightarrow{BM}=\left(\frac{13}{2};-2\right)\Rightarrow\overrightarrow{n_{BM}}=\left(2;\frac{13}{2}\right)\)
\(\Rightarrow BM:2\left(x+5\right)+\frac{13}{2}\left(y-1\right)=0\)
\(BM:2x+\frac{13}{2}y+\frac{7}{2}=0\)
d/ Gọi K là hình chiếu của B hạ xuống AC \(\Rightarrow BK\perp AC\)
\(\overrightarrow{n_{BK}}=\overrightarrow{AC}=\left(-1;-2\right)\)
\(\Rightarrow BK:-\left(x-2\right)-2y=0\)
\(BK:-x-2y+2=0\)
\(BK\cap AA'=\left\{H\right\}\Rightarrow\left\{{}\begin{matrix}-x-2y+2=0\\6x-3y-12=0\end{matrix}\right.\)
\(\Rightarrow H\left(2;0\right)\)
