Find the equation perpendicular to 10x – 9y + 2 = 0 and passing through point (3,3) is given by...

This is the answer given:

Eqn. of reqd line is of form 9x+10y+k = 0; k is to be determined.

(3,3)belongs to line=> 27+30+k = 0 or k = -57

9x+10y-57 = 0

Ans: C

I set up the point-slope :

y=10/9x + 2/9

m for the perp line is -9/10, but then how do I solve for the intercept?

y = -9/10x +c

but then I set

10y=-9x+10c

10y+9x=10c

c= y + 9/10x - that isn't right!