Question

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

Answer 1

y=1/3x-3

Answer 2

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

Answer 3

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 ...