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What is the equation of the line that is perpendicular to y = one-fifth x + 4 and that passes through (5,–4)?

On a coordinate plane, a line goes through (0, 4) and (5, 5). A point is at (5, negative 4).
y = negative 5 x minus 29
y = negative 5 x + 21
y = one-fifth x minus 4
y = one-fifth x minus 5

Respuesta :

Answer:

2nd option

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = [tex]\frac{1}{5}[/tex] x + 4 ← is in slope- intercept form

with slope m = [tex]\frac{1}{5}[/tex]

given a line with slope m then the slope of a line perpendicular to it is

[tex]m_{perpendicular}[/tex] = - [tex]\frac{1}{m}[/tex] = - [tex]\frac{1}{\frac{1}{5} }[/tex] = - 5 , then

y = - 5x + c ← is the partial equation

to find c substitute (5, - 4 ) into the partial equation

- 4 = - 25 + c ⇒ c = - 4 + 25 = 21

y = - 5x + 21 ← equation of perpendicular line

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