One more math question for you experts

[ARCHIVED THREAD] - One more math question for you experts

Posted: 2/21/2006 8:23:06 PM EDT

Write the equation of the line that passes through the given point and is parellel to the given line. Write the answer in slope-intercept form.

(3, 7), 5x - y = 10

Can anyone walk me through this one?

Posted: 2/21/2006 8:26:43 PM EDT

[#1]

First solve for Y to make things easier on yourself...therefore

y= 5x -10

Then, you have to put the line in point-slope form, which is:

y - y1 = m(x - x 1 )

Y1 = 7 and X1 = 3.

Thus.....

y - 7 = m(x - 3)

Now, figure out the slope and solve for y and you're in like foreskin.

Posted: 2/21/2006 8:27:18 PM EDT

[#2]

We begin by finding the slope of the line you must be parallel to:

5x-y=10

-y=10-5x

y=5x-10

so the slope is 5, crossing at (0,-10)

So, the equation you are looking for will be y=5x+A, where A needs to be solved for.

Since you know the line crosses through (3,7), you substitute for x and y, and get 7=5(3)+A, so A=-8

The equation of your line is Y=5X-8.

Posted: 2/21/2006 8:31:45 PM EDT

[#3]

Quoted:
First solve for Y to make things easier on yourself...therefore

y= 5x -10

Then, you have to put the line in point-slope form, which is:

y - y1 = m(x - x 1 )

Y1 = 7 and X1 = 3.

Thus.....

y - 7 = m(x - 3)

Now, figure out the slope and solve for y and you're in like foreskin.

I don't change the sign of 5x when I move it to the other side of the = sign? I had -y = -5 - 10.

Why don't we change the signs? This is obviously where I am screwing myself up.

Posted: 2/21/2006 8:32:31 PM EDT

[#4]

He changed all the signs at once. Notice the 10 went to -10.

I did it step by step in mine.

Posted: 2/21/2006 8:33:08 PM EDT

[#5]

Quoted:
Quoted:
First solve for Y to make things easier on yourself...therefore

y= 5x -10

Then, you have to put the line in point-slope form, which is:

y - y1 = m(x - x 1 )

Y1 = 7 and X1 = 3.

Thus.....

y - 7 = m(x - 3)

Now, figure out the slope and solve for y and you're in like foreskin.

I don't change the sign of 5x when I move it to the other side of the = sign? I had -y = -5 - 10.

Why don't we change the signs? This is obviously where I am screwing myself up.

If you had

5x - y = 10

for your original equation you can solve for it by subtracting -y (which is the same as adding y) and subtract 10. This leaves y = 5x - 10

Posted: 2/21/2006 8:34:21 PM EDT

[#6]

y+1 to the slant to (3, 7), 5x - y = 10 then add 4+3 and you get a big pile of SHEOT!!

Posted: 2/21/2006 8:34:40 PM EDT

[#7]

Quoted:
We begin by finding the slope of the line you must be parallel to:

5x-y=10

-y=10-5x

y=5x-10

so the slope is 5, crossing at (0,-10)

So, the equation you are looking for will be y=5x+A, where A needs to be solved for.

Since you know the line crosses through (3,7), you substitute for x and y, and get 7=5(3)+A, so A=-8

The equation of your line is Y=5X-8.

OK, how do I confirm that is parellel? Just graph it, or is there a way to check it in linear equation format?

I really appreciate the help folks! I think my brain is completely melted.

Posted: 2/21/2006 8:35:15 PM EDT

[#8]

If the slope is the same, the line is parallel.

Posted: 2/21/2006 8:35:37 PM EDT

[#9]

They both have a slope of 5, so by definition, they are parallel.

Posted: 2/21/2006 8:37:33 PM EDT

[#10]

Well DUH!

Told you my brain was freaking fried!

Posted: 2/21/2006 8:38:43 PM EDT

[#11]

Quoted:
Quoted:
First solve for Y to make things easier on yourself...therefore

y= 5x -10

Then, you have to put the line in point-slope form, which is:

y - y1 = m(x - x 1 )

Y1 = 7 and X1 = 3.

Thus.....

y - 7 = m(x - 3)

Now, figure out the slope and solve for y and you're in like foreskin.

I don't change the sign of 5x when I move it to the other side of the = sign? I had -y = -5 - 10.

Why don't we change the signs? This is obviously where I am screwing myself up.

when you have that "-y=-5x+10" or whenever you end up w/ that "-" in front of your unknown variable, just switch ALL the sights (take the opposite) or in mathimatical thought, multiply all the #s by a "-1"

golly, i miss simple math

Posted: 2/21/2006 8:39:36 PM EDT

[#12]

Quoted:
Quoted:
We begin by finding the slope of the line you must be parallel to:

5x-y=10

-y=10-5x

y=5x-10

so the slope is 5, crossing at (0,-10)

So, the equation you are looking for will be y=5x+A, where A needs to be solved for.

Since you know the line crosses through (3,7), you substitute for x and y, and get 7=5(3)+A, so A=-8

The equation of your line is Y=5X-8.

OK, how do I confirm that is parellel? Just graph it, or is there a way to check it in linear equation format?

I really appreciate the help folks! I think my brain is completely melted.

get a TI-83+ and graph it

Posted: 2/21/2006 8:39:48 PM EDT

[#13]

Quoted:

golly, i miss simple math

Amen!

Posted: 2/21/2006 8:42:08 PM EDT

[#14]

OK, so did I do this one right? This is perpendicular, not parallel

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

3x – 6y = 9

- 6y = -3x + 9

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

y = 1/2x + -3

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

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

y + 1 = 2x + 4

y + 1 – 1 = 2x + 4 – 1

y = 2x + 3

Posted: 2/21/2006 8:45:38 PM EDT

[#15]

Ah, shit. Now you had to go perpendicular on us!

Assuming that the slope of a line perpendicular to a line of slope m is 1/m, then yes, it seems you got it right.

I honestly don't remember the assumption, though....

ETA: Nope. The slope is -2. Negative reciprocal.

Your answer should be y = -2x + 3

Neat little app: Link

Posted: 2/21/2006 8:55:11 PM EDT

[#16]

Quoted:
Quoted:

golly, i miss simple math

Amen!

+1

Posted: 2/21/2006 9:18:57 PM EDT

[#17]

Thanks again for all the great help fellas! This has definately been a taxing class. I'll start asking more questions from now on - perhaps I can grasp some of this nonsense, as y'all explain it differently and it seems to make sense.

I tip one to your efforts in my appreciation

Posted: 2/21/2006 9:25:07 PM EDT

[#18]

Wait till you get to calculus!

Integration by parts is SO much fun!

[ARCHIVED THREAD] - One more math question for you experts

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