Algebra
posted by John .
Still need help on this one
Drawing a blank on this one, not sure where to start.
Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 12 and passing through (9, 5).

Very easy
Recall that the slopes of perpendicular lines have slopes that are negative reciprocals of each other, or in other words, the product of their slopes is 1
Your first line is x + 3y = 12
So the perpendicular line must be
3x  y = c , the only thing we don't know is the constant
But (9,5) lies on this new line
so 3(9)  (5) = c
c = 32
The equation must be 3x  y = 32 
I got something totally different 
x +3y = 12
x  x + 3y = 12  x
3y = x + 12
3y/3 = x/3 + 12/3
y = 3x + 4
y  5 = 1/3(x  9)
y + 5 = 1/3x  3
y + 5  5 = 1/3x  3  5
y = 1/3x  8 
no!
the slope of the given line was 1/3, the slope of your new line is +1/3
So they are not even perpendicular!!
from your line:
3y/3 = x/3 + 12/3 so far so good
now
y = (1/3)x + 4
so the new slope must be +3
y + 5 = 3(x  9)
y + 5 = 3x  27
y = 3x  32, which when rearranged is the same as 3x  y = 32
let me show you another approach
the slope of the given line was 1/3, 
(continued)
so the slope of the new line must be +3
then the new equation is y = 3x + b
but (9,5) lies on it
so 5 = (3)(9) + b
b = 32
for y = 3x  32 
Thank you so much.