Question

giúp em vs ạ

Answer 1

Chọn A

Answer 2

Gọi \(H\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AH}=\left(x+1;y-2\right)\\\overrightarrow{AB}=\left(-4;2\right)\\\overrightarrow{CH}=\left(x-2;y-4\right)\end{matrix}\right.\)

CH là đường cao hạ từ C \(\Rightarrow\left\{{}\begin{matrix}H\in AB\\CH\perp AB\end{matrix}\right.\)

 \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+1}{-4}=\dfrac{y-2}{2}\\-4\left(x-2\right)+2\left(y-4\right)=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{5}\\y=\dfrac{6}{5}\end{matrix}\right.\)

\(\Rightarrow H\left(\dfrac{3}{5};\dfrac{6}{5}\right)\)