Đáp án:
$\begin{array}{l}
2)\Delta //d;A \in \Delta \\
\Rightarrow \Delta :\left\{ \begin{array}{l}
x = 2t + 1\\
y = 4t - 3
\end{array} \right.\\
3)\\
d:3x - 2y + 1 = 0\\
\Rightarrow y = \frac{3}{2}x + \frac{1}{2}\\
Do:\Delta \bot d\\
\Rightarrow {k_\Delta } = \left( { - 1} \right):\frac{3}{2} = - \frac{2}{3}\\
\Rightarrow \Delta :y = - \frac{2}{3}x + b\\
Do:B\left( {3; - 1} \right) \in \Delta \\
\Rightarrow - 1 = - \frac{2}{3}.3 + b\\
\Rightarrow b = 1\\
Vậy\,\Delta :y = - \frac{2}{3}x + 1
\end{array}$
$\begin{array}{l}
4)M\left( { - 1 + 2t;2 + t} \right)\\
\Rightarrow AM = \sqrt {{{\left( { - 1 + 2t - 2} \right)}^2} + {{\left( {2 + t - 1} \right)}^2}} \\
\Rightarrow 10 = {\left( {2t - 3} \right)^2} + {\left( {t + 1} \right)^2}\\
\Rightarrow 5{t^2} - 12t + 2t + 10 = 10\\
\Rightarrow 5{t^2} - 10t = 0\\
\Rightarrow \left[ \begin{array}{l}
t = 0\\
t = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
M\left( { - 1;2} \right)\\
M\left( {3;4} \right)
\end{array} \right.
\end{array}$
