Đáp án:
a) \(PTTS:\left\{ \begin{array}{l}
x = 2 + 5t\\
y = 3 + 2t
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Do:\left( {d'} \right)//\left( d \right)\\
\to vtpt:\overrightarrow n = \left( {4; - 10} \right)\\
\to vtpt:\overrightarrow n = \left( {2; - 5} \right)\\
\to vtcp:\overrightarrow u = \left( {5;2} \right)\\
PTTQ:\left( {d'} \right):2\left( {x - 2} \right) - 5\left( {y - 3} \right) = 0\\
\to 2x - 5y + 11 = 0\\
PTTS:\left\{ \begin{array}{l}
x = 2 + 5t\\
y = 3 + 2t
\end{array} \right.\\
PTCT:\dfrac{{x - 2}}{5} = \dfrac{{y - 3}}{2}\\
d)Do:\left( {d'} \right)//\left( d \right)\\
\to vtcp:\overrightarrow u = \left( { - 2;4} \right)\\
\to vtcp:\overrightarrow u = \left( { - 1;2} \right)\\
\to vtpt:\overrightarrow n = \left( {2;1} \right)\\
\to PTTQ:\left( {d'} \right):2\left( {x - 2} \right) + y + 3 = 0\\
\to 2x + y - 1 = 0\\
PTTS:\left\{ \begin{array}{l}
x = 2 - t\\
y = - 3 + 2t
\end{array} \right.\\
PTCT:\dfrac{{x - 2}}{{ - 1}} = \dfrac{{y + 3}}{2}\\
b)Do:\left( {d'} \right)//\left( d \right)\\
\to vtpt:\overrightarrow n = \left( {0;1} \right)\\
\to vtcp:\overrightarrow u = \left( {1;0} \right)\\
\to PTTQ:\left( {d'} \right):0.\left( {x + 1} \right) + y - 2 = 0\\
\to y - 2 = 0\\
PTTS:\left\{ \begin{array}{l}
x = - 1 + t\\
y = 2
\end{array} \right.\\
e)Do:\left( {d'} \right)//\left( d \right)\\
\to vtcp:\overrightarrow u = \left( {3; - 2} \right)\\
\to vtpt:\overrightarrow n = \left( {2;3} \right)\\
\to PTTQ:\left( {d'} \right):2x + 3\left( {y - 3} \right) = 0\\
\to 2x + 3y - 9 = 0\\
PTTS:\left\{ \begin{array}{l}
x = 3t\\
y = 3 - 2t
\end{array} \right.\\
PTCT:\dfrac{x}{3} = \dfrac{{y - 3}}{{ - 2}}
\end{array}\)
\(\begin{array}{l}
c)Do:\left( {d'} \right)//\left( d \right)\\
\to vtpt:\overrightarrow n = \left( {1;0} \right)\\
\to vtcp:\overrightarrow u = \left( {0;1} \right)\\
\to PTTQ:\left( {d'} \right):x - 4 = 0\\
PTTS:\left\{ \begin{array}{l}
x = 4\\
y = 3 + t
\end{array} \right.
\end{array}\)
