Đáp án:
\(N(\frac{2}{3},\frac{{13}}{3})\)
\(M(\frac{7}{3},3)\)
Giải thích các bước giải:
\(\begin{array}{l}
{S_{ANMC}} = 8{S_{BMN}}\\
{S_{ANMC}} + {S_{BMN}} = {S_{ABC}}\\
\to {S_{BMN}} = \frac{{{S_{ABC}}}}{9}
\end{array}\)
MN//AC -> áp dụng Ta-lét:
\(\frac{{{S_{BMN}}}}{{{S_{ABC}}}} = {\left( {\frac{{BN}}{{AB}}} \right)^2} = {\left( {\frac{{BM}}{{BC}}} \right)^2}\)
\(\begin{array}{l}
\to {\left( {\frac{{BN}}{{AB}}} \right)^2} = {\left( {\frac{{BM}}{{BC}}} \right)^2} = \frac{1}{9} \leftrightarrow \frac{{BN}}{{AB}} = \frac{{BM}}{{BC}} = \frac{1}{3}\\
\to 3\overrightarrow {BN} = \overrightarrow {BA} \leftrightarrow \left\{ \begin{array}{l}
3({x_N} - 2) = - 2 - 2\\
3({y_N} - 5) = 3 - 5
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
{x_N} = \frac{2}{3}\\
{y_N} = \frac{{13}}{3}
\end{array} \right. \leftrightarrow N(\frac{2}{3},\frac{{13}}{3})
\end{array}\)
\( \to 3\overrightarrow {BM} = \overrightarrow {BC} \leftrightarrow \left\{ \begin{array}{l}
3({x_M} - 2) = 3 - 2\\
3({y_M} - 5) = - 1 - 5
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
{x_M} = \frac{7}{3}\\
{y_M} = 3
\end{array} \right. \leftrightarrow M(\frac{7}{3},3)\)
