Đáp án:
\(P(0;\,\,0)\)
Giải thích các bước giải:
Ta có: \(P \in Oy \Rightarrow P\left( {0;\,\,a} \right).\)
\( \Rightarrow \left\{ \begin{array}{l}\overrightarrow {PM} = \left( {1;\,\,2 - a} \right) \Rightarrow P{M^2} = 1 + {\left( {2 - a} \right)^2}\\\overrightarrow {PN} = \left( { - 2;\,\,1 - a} \right) \Rightarrow P{N^2} = 4 + {\left( {1 - a} \right)^2}\end{array} \right.\)
\(\Delta MPN\) vuông cân tại \(P \Leftrightarrow \left\{ \begin{array}{l}PN \bot PM\\PN = PM\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\overrightarrow {PN} .\overrightarrow {PM} = 0\\P{N^2} = P{M^2}\end{array} \right.\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}1.\left( { - 2} \right) + \left( {2 - a} \right)\left( {1 - a} \right) = 0\\1 + {\left( {2 - a} \right)^2} = 4 + {\left( {1 - a} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 2 + 2 - 3a + {a^2} = 0\\{a^2} - 4a + 5 = {a^2} - 2a + 5\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a^2} - 3a = 0\\2a = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a\left( {a - 3} \right) = 0\\a = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}a = 0\\a = 3\end{array} \right.\\a = 0\end{array} \right. \Leftrightarrow a = 0 \Rightarrow P\left( {0;\,\,0} \right)\end{array}\)
