Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1).

y = one thirdx − 3

y = one thirdx + 17

y = −3x − 3

y = −3x + 17

Question

Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1). 

 y = one thirdx − 3 

 y = one thirdx + 17 

 y = −3x − 3 

 y = −3x + 17

anonymous · Answer

Ok..Two lines are perpendicular if their slopes are negative reciprocals of each other

anonymous · Answer

3x+y=7 can be written as y=-3x+7...so the slope of this line is -3 and the slope of our perpendicular line is -(1/-3)=1/3.

anonymous · Answer

it passes through (6,-1) so (y-y1)=m(x-x1) --->(y--1)=1/3(x-6) =y+1=(1/3)x-2

anonymous · Answer

y=(1/3)x-3

anonymous · Answer

@BSANDSTORM