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Write an equation of the line that is perpendicular to 3x + 9y = 7 and passes through the point (6, 4). A) y = 3x - 14 B) y = 3x + 4 C) y = -3x + 16 D) y = 1 3 x - 2

Respuesta :

calculista
we have that
3x + 9y = 7-------> 9y=7-3x----> y=(7/9)-(1/3)*x
and 
point (6, 4)

we know that
if two lines are perpendicular so
m1*m2=-1

step 1
find the slope m2
m2=-1/m1

y=(7/9)-(1/3)*x---------> slope m1=-1/3
so
m2=3

step 2
find the equation of a line
with m2=3 and the point  (6, 4)

y-y1=m*(x-x1)------> y-4=3*(x-6)----> y=3x-18+4----> y=3x-14

the answer is the option
A) y = 3x - 14

see the attached figure
Ver imagen calculista
abbiosborne39

Answer:

A

Step-by-step explanation:

Solution: y = 3x - 26. First put the original equation in slope-intercept form: y= 1/3x+ 7/9

. Perpendicular lines have slopes that are opposite reciprocals of each other. Therefore, the slope of the line is 3. When we solve for the y-intercept, the result is -14.