Trong mặt phẳng tọa độ Oxy, cho tam giác ABC có A(3;4), B(2;1), C(-1; -2

* Hướng dẫn giải

Đáp án C

Dễ thấy $\frac{S_{\Delta ABC}}{S_{\Delta ABM}}=4\iff \frac{BC}{BM}=4\iff \left[\begin{array}{c}\overrightarrow{BC}=4\overrightarrow{BM}\\ \overrightarrow{BC}=-4\overrightarrow{BM}\end{array}\right.$

TH1: $\overrightarrow{BC}=4\overrightarrow{BM}\Rightarrow \overrightarrow{BM}=\frac{1}{4}\overrightarrow{BC}$ 

thì $\left\{\begin{array}{c}x-2=-\frac{3}{4}\\ y-1=-\frac{3}{4}\end{array}\right.\iff \left\{\begin{array}{c}x=\frac{5}{4}\\ y=\frac{1}{4}\end{array}\right.\Rightarrow x.y=\frac{5}{16}$

TH2: $\overrightarrow{BC}=-4\overrightarrow{BM}\Rightarrow \overrightarrow{BM}=-\frac{1}{4}\overrightarrow{BC}$

thì $\left\{\begin{array}{c}x-2=\frac{3}{4}\\ y-1=\frac{3}{4}\end{array}\right.\iff \left\{\begin{array}{c}x=\frac{11}{4}\\ y=\frac{7}{4}\end{array}\right.\Rightarrow x.y=\frac{77}{16}$