# Algebra

posted by on .

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).

• Algebra - ,

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

• Algebra - ,

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

• Algebra - ,

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!!

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,

• Algebra - ,

(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

• Algebra - ,

Thank you so much.