$\begin{array}{l} M \in y'Oy \Rightarrow M\left( {0;\,\,m} \right)\\ Pt\,\,AB:\,\,\,\frac{{x + 2}}{{4 + 2}} = \frac{{y + 3}}{{7 + 3}}\\ \Leftrightarrow 10\left( {x + 2} \right) = 6\left( {y + 3} \right)\\ \Leftrightarrow 5x + 10 = 3y + 9\\ \Leftrightarrow 5x - 3y + 1 = 0\\ M,\,\,A,\,\,B\,\,\,thang\,\,hang\\ \Rightarrow M \in AB\\ \Rightarrow 5.0 - 3.m + 1 = 0\\ \Leftrightarrow 3m = 1\\ \Leftrightarrow m = \frac{1}{3}\\ \Rightarrow M\left( {0;\,\,\,\frac{1}{3}} \right). \end{array}$

Đáp án:

\(M\left( {0;\frac{1}{3}} \right)\).

Giải thích các bước giải: PT AB: \(\begin{array}{l}\frac{{x + 2}}{{4 + 2}} = \frac{{y + 3}}{{7 + 3}}\\ \Leftrightarrow \frac{{x + 2}}{6} = \frac{{y + 3}}{{10}}\\ \Leftrightarrow 5\left( {x + 2} \right) = 3\left( {y + 3} \right)\\ \Leftrightarrow 5x - 3y + 1 = 0\end{array}\) \(M = AB \cap Oy\). Cho \(x = 0 \Leftrightarrow - 3y + 1 = 0 \Leftrightarrow y = \frac{1}{3}\). Vậy \(M\left( {0;\frac{1}{3}} \right)\).