What is the equation of the line that is perpendicular to 9x+3y=36 and passes through the point (6,2)?

9x+3y=36

3y=-9x+36

y=-3x+36

For lines to be perpendicular their slopes must be negative reciprocals of each other...mathematically:

m1*m2=-1 and in this case for a line to be perpendicular to y=-3x+36 that line must have a slope of:

-3m=-1, m=1/3 so

y=x/3+b, and using point (6,2) we can solve for b, the y-intercept...

2=6/3+b

2=3+b

b=-1 so the perpendicular line passing through the point (6,2) is:

y=x/3-1 or more neatly...

y=(x-3)/3