Trong mặt phẳng tọa độ \(Oxy\), cho tam giác \(ABC\) có \(A\left( {3\,;\,4} \right)\), \(B\left( {2\,;\,1} \right)\), \(C\left( { - 1\,;\, - 2} \right)\). Gọi \(M\left( {x\,;\,y} \right)\) là điểm trên đường thẳng \(BC\) sao cho \({S_{\Delta ABC}} = 4{S_{\Delta ABM}}\). Tính \(P = x.y\).

Dễ thấy \(\dfrac{{{S_{\Delta ABC}}}}{{{S_{\Delta ABM}}}} = 4\)\( \Leftrightarrow \dfrac{{BC}}{{BM}} = 4\)\( \Leftrightarrow \left[ \begin{array}{l}\overrightarrow {BC}  = 4\overrightarrow {BM} \\\overrightarrow {BC}  =  - 4\overrightarrow {BM} \end{array} \right.\).

TH1: \(\overrightarrow {BC}  = 4\overrightarrow {BM} \) thì: \(\left\{ \begin{array}{l}x - 2 =  - \dfrac{3}{4}\\y - 1 =  - \dfrac{3}{4}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{5}{4}\\y = \dfrac{1}{4}\end{array} \right.\)\( \Rightarrow x.y = \dfrac{5}{{16}}\).

TH2: \(\overrightarrow {BC}  =  - 4\overrightarrow {BM} \) thì: \(\left\{ \begin{array}{l}x - 2 = \dfrac{3}{4}\\y - 1 = \dfrac{3}{4}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{{11}}{4}\\y = \dfrac{7}{4}\end{array} \right.\)\( \Rightarrow x.y = \dfrac{{77}}{{16}}\).