Đáp án:
\[\left[ \begin{array}{l}
M\left( { - 3; - 2} \right)\\
M\left( { - 5;4} \right)
\end{array} \right.\]
Giải thích các bước giải:
Gọi \(M\left( {a;b} \right)\) là điểm thỏa mãn tam giác ABM vuông cân tại A
Suy ra \(\left\{ \begin{array}{l}
AM \bot AB\\
AM = AB
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\overrightarrow {AM} .\overrightarrow {AB} = \overrightarrow 0 \\
AM = AB
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\overrightarrow {AB} \left( {3;1} \right)\\
\overrightarrow {AM} \left( {a + 4;b - 1} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
AB = \sqrt {10} \\
AM = \sqrt {{{\left( {a + 4} \right)}^2} + {{\left( {b - 1} \right)}^2}}
\end{array} \right.\\
\left\{ \begin{array}{l}
\overrightarrow {AB.} \overrightarrow {AM} = 0\\
AB = AM
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3\left( {a + 4} \right) + 1.\left( {b - 1} \right) = 0\\
10 = {\left( {a + 4} \right)^2} + {\left( {b - 1} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {b - 1} \right) = - 3\left( {a + 4} \right)\\
{\left( {a + 4} \right)^2} + {\left[ { - 3\left( {a + 4} \right)} \right]^2} = 10
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a + 4 = 1\\
b - 1 = - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
a + 4 = - 1\\
b - 1 = 3
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = - 3\\
b = - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
a = - 5\\
b = 4
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
M\left( { - 3; - 2} \right)\\
M\left( { - 5;4} \right)
\end{array} \right.
\end{array}\)
