În reperul cartezian \( xOy \) se consideră dreapta \( d \) de ecuație \( \color{fuchsia}a \color{grey}x + \color{darkmagenta}b \color{grey}y + c = 0 \) și punctul \( A(\color{blue}x_A \color{grey}, \color{orange}y_A \color{grey}) \).
Se va determina ecuația dreptei \( d^{'} \), paralela dusă prin punctul \( A \) la dreapta \( d \).

   \( d \parallel d^{'} \implies \color{red}m \color{grey} = \color{red}m^{'} \),
adică, fiind paralele, cele două drepte au pantele egale.

   trebuie aflată mai întâi panta dreptei \( d \)
\( d \) : \( \color{fuchsia}a \color{grey}x + \color{darkmagenta}b \color{grey}y + c = 0 \)
\( d \) : \( \color{darkmagenta}b \color{grey}y = - \color{fuchsia}a \color{grey}x - c \)
\( d \) : \( \displaystyle y = \frac{- \color{fuchsia}a}{\color{darkmagenta}b} \color{grey}x + \frac{- c}{\color{darkmagenta}b} \)

   pantele celor două drepte sunt
\( \displaystyle \color{red}m \color{grey} = \color{red}m^{'} \color{grey} = \frac{- \color{red}a}{\color{red}b}\)
   ecuația dreptei \( d^{'} \) este:
\( d^{'} \) : \( y - \color{orange}y_A \color{grey} = \color{red}m^{'} \color{grey}( x - \color{blue}x_A \color{grey}) \)
\( d^{'} \) : \( \displaystyle y - \color{orange}y_A \color{grey} = \frac{- \color{red}a}{\color{red}b} \color{grey}( x - \color{blue}x_A \color{grey}) \),
   de aici, efectuând calculele, se poate afla ecuația dreptei \( d^{'} \) în forma explicită sau forma generală:

forma explicită:
\( d^{'} \) : \( \displaystyle y - \color{orange}y_A \color{grey} = \frac{- \color{red}a}{\color{red}b} \color{grey}( x - \color{blue}x_A \color{grey}) \)

\( d^{'} \) : \( \displaystyle y - \color{orange}y_A \color{grey} = \frac{- \color{red}a}{\color{red}b} \color{grey} x + \frac{\color{red}a}{\color{red}b} \color{blue}x_A \color{grey} \)

\( d^{'} \) : \( \displaystyle y = \frac{- \color{red}a}{\color{red}b} \color{grey} x + \frac{\color{red}a}{\color{red}b} \color{blue}x_A \color{grey} + \color{orange}y_A\)

forma generală:
\( d^{'} \) : \( \displaystyle y - \color{orange}y_A \color{grey} = \frac{- \color{red}a}{\color{red}b} \color{grey}( x - \color{blue}x_A \color{grey}) \rvert \cdot \color{red}b\)

\( d^{'} \) : \( \displaystyle \color{red}b \color{grey} (y - \color{orange}y_A \color{grey}) = - \color{red}a \color{grey}( x - \color{blue}x_A \color{grey}) \)

\( d^{'} \) : \( \displaystyle \color{red}b \color{grey}y - \color{red}b \color{orange}y_A \color{grey} = - \color{red}a \color{grey}x + \color{red}a \color{blue}x_A \color{grey} \)

\( d^{'} \) : \( \displaystyle \color{red}a \color{grey}x + \color{red}b \color{grey}y - \color{red}b \color{orange}y_A \color{grey} - \color{red}a \color{blue}x_A \color{grey} = 0\).

---

În reperul cartezian \( xOy \) se consideră dreapta \( d \) de ecuație \( \color{fuchsia}5 \color{grey}x + \color{darkmagenta}2 \color{grey}y + 1 = 0 \) și punctul \( A(\color{blue}4 \color{grey}, \color{orange}-3 \color{grey}) \).
Se va determina ecuația dreptei \( d^{'} \), paralela dusă prin punctul \( A \) la dreapta \( d \).

   \( d \parallel d^{'} \implies \color{red}m \color{grey} = \color{red}m^{'} \),
adică, fiind paralele, cele două drepte au pantele egale.

   Mai întâi trebuie determinată panta dreptei \( d \)

\( d \) : \( \color{fuchsia}5 \color{grey}x + \color{darkmagenta}2 \color{grey}y + 1 = 0 \)
\( d \) : \( \color{darkmagenta}2 \color{grey}y = \color{fuchsia}-5 \color{grey}x - 1 \)
\( d \) : \( \displaystyle y = \frac{ \color{fuchsia}-5}{\color{darkmagenta}2} \color{grey}x + \frac{- 1}{\color{darkmagenta}2} \)

   pantele celor două drepte sunt
\( \displaystyle \color{red}m \color{grey} = \color{red}m^{'} \color{grey} = \frac{ \color{red}-5}{\color{red}2}\)

   ecuația dreptei \( d^{'} \) este:
\( d^{'} \) : \( y - \color{orange}y_A \color{grey} = \color{red}m^{'} \color{grey}( x - \color{blue}x_A \color{grey}) \)
\( d^{'} \) : \( \displaystyle y - (\color{orange}-3 \color{grey}) = \frac{- \color{red}5}{\color{red}2} \color{grey}( x - \color{blue}4 \color{grey}) \),

   de aici, efectuând calculele, se poate afla ecuația dreptei \( d^{'} \) în forma explicită sau forma generală:

forma explicită:
\( d^{'} \) : \( \displaystyle y - (\color{orange}-3 \color{grey}) = \frac{\color{red}-5}{\color{red}2} \color{grey}( x - \color{blue}4 \color{grey}) \)

\( d^{'} \) : \( \displaystyle y + \color{orange}3 \color{grey} = \frac{\color{red}-5}{\color{red}2} \color{grey} x + \frac{\color{red}5}{\color{red}2} \cdot \color{blue}4 \color{grey} \)

\( d^{'} \) : \( \displaystyle y = \frac{\color{red}-5}{\color{red}2} \color{grey} x + \frac{\color{red}5}{\color{red}2} \cdot \color{blue}4 \color{grey} - \color{orange}3 \color{grey} \)

\( d^{'} \) : \( \displaystyle y = \frac{\color{red}-5}{\color{red}2} \color{grey} x + 5 \cdot 2 - 3 \)

\( d^{'} \) : \( \displaystyle y = \frac{\color{red}-5}{\color{red}2} \color{grey} x + 7 \)


forma generală:
\( d^{'} \) : \( \displaystyle y - (\color{orange}-3 \color{grey}) = \frac{\color{red}-5}{\color{red}2} \color{grey}( x - \color{blue}4 \color{grey}) \rvert \cdot \color{red}2 \)

\( d^{'} \) : \( \displaystyle \color{red}2 \color{grey} (y + \color{orange}3 \color{grey}) = \color{red}-5 \color{grey}( x - \color{blue}4 \color{grey}) \)
\( d^{'} \) : \( \displaystyle 2y + 6 = - 5x + 20 \)
\( d^{'} \) : \( \displaystyle 5x + 2y + 6 - 20 = 0 \)
\( d^{'} \) : \( \displaystyle 5x + 2y - 14 = 0 \).

---