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November 22, 2014

November 22, 2014

Posted by **John** on Monday, October 15, 2007 at 10:39pm.

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 -
**Reiny**, Monday, October 15, 2007 at 10:51pmVery 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 -
**John**, Monday, October 15, 2007 at 11:30pmI 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 -
**Reiny**, Monday, October 15, 2007 at 11:40pmno!

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,

- Algebra -

- Algebra -
- Algebra -
**Reiny**, Monday, October 15, 2007 at 11:43pm(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 -
**John**, Tuesday, October 16, 2007 at 9:04amThank you so much.

- Algebra -

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