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aliya0001 [1]
3 years ago
11

Encuentra la forma de la ecuacion de la recta que pasa por el punto P(-1,5) y que es perpendicular a la recta 2x + 3y - 8= 0

Mathematics
2 answers:
finlep [7]3 years ago
4 0

Responder:

2y-3x = 12

Explicación paso a paso:

La forma estándar de ecuación de una línea se expresa como;

y = mx + c

m es la pendiente

c es la intersección

Primero necesitamos la pendiente de la ecuación conocida;

Dada la ecuación 2x ​​+ 3y - 8 = 0

Vuelva a escribir en forma estándar;

2x + 3y = 8

3y = -2x + 8

y = -2x / 3 + 8/3

Por lo tanto, la pendiente de la línea dada es -2/3

Dado que la línea desconocida es perpendicular a la línea dada, el producto de sus pendientes será -1;

mM = -1

M = -1 (-2/3)

M = 3/2

La pendiente de la línea desconocida es 3/2

Obtener la intersección:

Sustituya m = 3/2 y el punto (-1, 5) en la expresión y = mx + cy obtenga c como se muestra;

5 = 3/2 (-1) + c

5 = -3/2 + c

c = 5 + 3/2

c = (10 + 3) / 2

c = 13/2

Obtenga la ecuación requerida;

Dado que y = mx + c

y = 3x / 2 + 13/2

multiplicar por 2;

2y = 3x + 12

2y-3x = 12

De ahí que la forma de la ecuación de la recta que pasa por el punto P (-1,5) y que es perpendicular a la recta 2x + 3y - 8 = 0 es 2y-3x = 12

Dmitrij [34]3 years ago
3 0

Answer:

The answer is A, C, E

Step-by-step explanation:

I took the test

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