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

Question

anonymous · Answer

y=1/3x-3

anonymous · Answer

3x + y = 7
y=-3x+7  here slope m = -3  
the line perpendicular has slope m=-1/m=-1/(-3)=1/3
now at point(6,-1)=(x1,y1)
from y-y1=m(x-x1) we get
         y-(-1)=(1/3)(x-6) 
         y+1   =  (1/3)x -2
               y= (1/3)x -2-1=(1/3)x-3

amistre64 · Answer

SWAP COEFFS AND NEGATE ONE Of em ..... bad pinky!!

3x+y = 7

x-3y = 7 , that sevens from something other than what we want so delete in and find a new constant

x-3y = c

the given point has to fit ...