Trong mặt phẳng tọa độ Oxy, cho tam giác ABC có A(4; 3); B(2; 7)

* Hướng dẫn giải

Gọi $A\text{'}\left(x;y\right)$. Ta có $\left\{\begin{array}{l}\overrightarrow{AA\text{'}}=\left(x-4;y-3\right)\\ \overrightarrow{BC}=\left(-5;-15\right)\\ \overrightarrow{BA\text{'}}=\left(x-2;y-7\right)\end{array}\right..$

Từ giả thiết, ta có $\left\{\begin{array}{l}AA\text{'}\perp BC\\ B,A\text{'},C\text{ thang hang}\end{array}\right.\iff \left\{\begin{array}{ll}\overrightarrow{AA\text{'}}.\overrightarrow{BC}=0& \left(1\right)\\ \overrightarrow{BA\text{'}}=k\overrightarrow{BC}& \left(2\right)\end{array}\right..$

 $\left(1\right)\iff -5\left(x-4\right)-15\left(y-3\right)=0\iff x+3y=13.$ 

 $\left(2\right)\iff \frac{x-2}{-5}=\frac{y-7}{-15}\iff 3x-y=-1.$

Giải hệ $\left\{\begin{array}{l}x+3y=13\\ 3x-y=-1\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ y=4\end{array}\right.\Rightarrow A\text{'}\left(1;4\right).$ 

Chọn C.