Đáp án:

f) \(PTCT:\dfrac{x}{5} = \dfrac{y}{{ - 2}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)Do:vtpt:\overrightarrow n  = \left( {5; - 1} \right) \to vtcp:\overrightarrow u  = \left( {1;5} \right)\\
PTTQ:5\left( {x + 2} \right) - \left( {y - 3} \right) = 0\\
 \to 5x - y + 13 = 0\\
PTTS:\left\{ \begin{array}{l}
x =  - 2 + t\\
y = 3 + 5t
\end{array} \right.\\
PTCT:\dfrac{{x + 2}}{1} = \dfrac{{y - 3}}{5}\\
d)Do:vtpt:\overrightarrow n  = \left( {5;0} \right) \to vtcp:\overrightarrow u  = \left( {0;5} \right)\\
PTTQ:5\left( {x - 1} \right) + 0.\left( {y - 2} \right) = 0\\
 \to 5x - 5 = 0\\
PTTS:\left\{ \begin{array}{l}
x = 1\\
y = 2 + 5t
\end{array} \right.\\
b)Do:vtpt:\overrightarrow n  = \left( { - 2;3} \right) \to vtcp:\overrightarrow u  = \left( {3;2} \right)\\
PTTQ: - 2\left( {x + 1} \right) + 3\left( {y - 2} \right) = 0\\
 \to  - 2x + 3y - 8 = 0\\
PTTS:\left\{ \begin{array}{l}
x =  - 1 + 3t\\
y = 2 + 2t
\end{array} \right.\\
PTCT:\dfrac{{x + 1}}{3} = \dfrac{{y - 2}}{2}\\
e)Do:vtpt:\overrightarrow n  = \left( {0;3} \right) \to vtcp:\overrightarrow u  = \left( {3;0} \right)\\
PTTQ:0\left( {x - 7} \right) + 3\left( {y + 3} \right) = 0\\
 \to 3y + 9 = 0\\
PTTS:\left\{ \begin{array}{l}
x = 7 + 3t\\
y =  - 3
\end{array} \right.\\
c)Do:vtpt:\overrightarrow n  = \left( { - 2; - 5} \right) \to vtcp:\overrightarrow u  = \left( {5; - 2} \right)\\
PTTQ: - 2\left( {x - 3} \right) - 5\left( {y + 1} \right) = 0\\
 \to  - 2x - 5y + 1 = 0\\
PTTS:\left\{ \begin{array}{l}
x = 3 + 5t\\
y =  - 1 - 2t
\end{array} \right.\\
PTCT:\dfrac{{x - 3}}{5} = \dfrac{{y + 1}}{{ - 2}}\\
f)Do:vtpt:\overrightarrow n  = \left( {2;5} \right) \to vtcp:\overrightarrow u  = \left( {5; - 2} \right)\\
PTTQ:2x + 5y = 0\\
PTTS:\left\{ \begin{array}{l}
x = 5t\\
y =  - 2t
\end{array} \right.\\
PTCT:\dfrac{x}{5} = \dfrac{y}{{ - 2}}
\end{array}\)