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Posted: 2/21/2006 6:01:33 PM EDT
Folks, I am having a helluva time with this question.  I don't think I am doing it right.  Can anyone help?

Find the equation, in standard form, of the line perpendicular to 3x – 6y = 9 and passing through (-2, -1).


This is what I have so far, but it can't be right.  

3x – 6y = 9

- 6y = 3x + 9

-6y/-6 = 3x/-6 + 9/-6

y = (-1/2)x + -3

y = 1x + -3  

How do I turn (-2, -1) into a linear equation like 3x - 6y = 9 to show the slopes as co-efficient?

Thanks folks!
Link Posted: 2/21/2006 6:07:36 PM EDT
[#1]

Quoted:
Folks, I am having a helluva time with this question.  I don't think I am doing it right.  Can anyone help?

Find the equation, in standard form, of the line perpendicular to 3x – 6y = 9 and passing through (-2, -1).


This is what I have so far, but it can't be right.  

3x – 6y = 9

- 6y = 3x + 9

-6y/-6 = 3x/-6 + 9/-6

y = (-1/2)x + -3

y = 1x + -3  

How do I turn (-2, -1) into a linear equation like 3x - 6y = 9 to show the slopes as co-efficient?

Thanks folks!




3x – 6y = 9

is equivalent to 6y = 3x - 9 equivalent to y = (1/2)x - 3/2
so m = 1/2
this is now in the form of y=mx + b where m is the slope, and b is the y intercept.

a line with a Perpendicular slope to m would have a slope of -1/m (m <> 0)
let n = -1/m = -2
IF m is 0 then the perpendicular is parallel to the y axis.

Since the new line also must pass through the point (-2, -1) we know that
from y = nx + b we get  -1 = -2n + b

but since  n = -2

we get -1 = -2*(-2) + b
solving for b we get -1 = 4 + b or b = -4 + (-1) or b = -5

therefore

our new equation is: y = -2x - 5 which is perpendicular to: y = (1/2)x - 3/2  or 3x – 6y = 9
and our new equation passes through point (-2, -1)





Link Posted: 2/21/2006 6:08:24 PM EDT
[#2]
What's the use?
Link Posted: 2/21/2006 6:09:56 PM EDT
[#3]
use your equation y = (-1/2)x + -3. (-1/2) equals your slope (m).  since perpendicular lines have the opposite slope, the slope you have to use is 2. So then you use the equation (y-y1) = m(x-x1). you plug in the points that you are given (-2, -1) for y1 and x1, and plug in 2 for m. and solve for y.

(y - y1) = m(x - x1)

(y - (-1) ) = (2)(x - (-2) )

y + 1 = 2x + 4

y = 2x + 3 should be your answer

someone check this to make sure I'm right
Link Posted: 2/21/2006 6:15:53 PM EDT
[#4]
second equation is wrong, check ur negatives.
Link Posted: 2/21/2006 6:19:44 PM EDT
[#5]

Quoted:
second equation is wrong, check ur negatives.



how do you figure? the double negatives make positive
Link Posted: 2/21/2006 6:23:39 PM EDT
[#6]
I'd help you, but I was never any good at Spanish.
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