Tam giác ABC có C(–2; –4), trọng tâm G(0; 4), trung điểm cạnh BC

* Hướng dẫn giải

$\left\{\begin{array}{c}{x}_M=\frac{x_B+\text{}{x}_C}{2}\\ y_M=\frac{y_B+\text{}{y}_C}{2}\end{array}\right.\iff \left\{\begin{array}{c}{x}_B=2x_M-x_C=2.2-(-2)=6\\ y_B=2y_M-y_C=2.0-(-4)=4\end{array}\right.\Rightarrow B(6;4)$

Do G là trọng tâm tam giác ABC nên:

$\left\{\begin{array}{c}{x}_G=\frac{x_A+\text{}{x}_B+\text{}{x}_C}{3}\\ y_G=\frac{y_A+\text{}{y}_B+\text{}{y}_C}{3}\end{array}\right.\iff \left\{\begin{array}{c}{x}_A=3x_G-x_B-x_C=3.0-6-(-2)=-4\\ y_A=3y_G-y_B-y_C=3.4-4-(-4)=12\end{array}\right.\Rightarrow A(-4;12)$

Đáp án C