Đáp án:
$\begin{array}{l}
a)VTCP:\overrightarrow u \left( {3; - 4} \right);M\left( {2;3} \right)\\
\Rightarrow PTTS:\left\{ \begin{array}{l}
x = 2 + 3t\\
y = 3 - 4t
\end{array} \right.\\
b)VTPT:\overrightarrow n \left( {1;5} \right)\\
\Rightarrow VTCP:\overrightarrow u \left( {5; - 1} \right)\\
\Rightarrow PTTS:\left\{ \begin{array}{l}
x = 2 + 5t\\
y = 3 - t
\end{array} \right.\\
c)k = 3\\
\Rightarrow y = 3.x - 3\\
\Rightarrow 3.x - y - 3 = 0\\
\Rightarrow VTCP:\overrightarrow u = \left( {3; - 1} \right)\\
\Rightarrow PTTS:\left\{ \begin{array}{l}
x = 2 + 3t\\
y = 3 - t
\end{array} \right.\\
e)\Delta :2x + 3y + 1 = 0\\
\Rightarrow \left\{ \begin{array}{l}
VTCP:\overrightarrow u = \left( {2;3} \right)\\
M\left( {1; - 1} \right) \in \Delta
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 1 + 2t\\
y = - 1 + 3t
\end{array} \right.
\end{array}$
