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