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Hallar la ecuación de la recta paralela a \( y=2 x-5 \) que pasa por el punto \( P(-3 ; 3) \)

Ask by Cruz Hodges. in Argentina
Feb 03,2025

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La ecuación de la recta paralela a \( y = 2x - 5 \) que pasa por el punto \( P(-3, 3) \) es \( y = 2x + 9 \).

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Para hallar la ecuación de la recta paralela a \( y=2x-5 \), debemos entender que las rectas paralelas tienen la misma pendiente. En este caso, la pendiente de la recta original es \( 2 \). Usando la forma punto-pendiente de la ecuación de una recta, que es \( y - y_1 = m(x - x_1) \), donde \( m \) es la pendiente y \( (x_1, y_1) \) es el punto dado \( P(-3, 3) \), podemos sustituir: \( y - 3 = 2(x + 3) \). Desarrollando esto, obtenemos: \( y - 3 = 2x + 6 \) \( y = 2x + 9 \). Así que la ecuación de la recta paralela que pasa por el punto \( P(-3, 3) \) es \( y = 2x + 9 \).

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