Tìm toạ độ điểm D.

Hướng dẫn giải

Gọi D(x; y), do \(\overrightarrow {OD} = \frac{2}{3}\overrightarrow {OA} + \frac{1}{3}\overrightarrow {OB} \), suy ra: \(\left\{ \begin{array}{l}x = \frac{2}{3}{x_A} + \frac{1}{3}{x_B}\\y = \frac{2}{3}{y_A} + \frac{1}{3}{y_B}\end{array} \right.\).

\( \Rightarrow \left\{ \begin{array}{l}x = \frac{2}{3}.3 + \frac{1}{3}.8 = \frac{{14}}{3}\\y = \frac{2}{3}.4 + \frac{1}{3}.6 = \frac{{14}}{3}\end{array} \right.\).

Vậy \(D\left( {\frac{{14}}{3};\,\,\frac{{14}}{3}} \right)\).