Simple eqN of a line question

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July 30th 2006, 04:35 PM
NineZeroFive

Simple eqN of a line question

Hey,

I have this question thats been bugging me for a few days now.

$3y-15=2x-8$

Whats the answer if we put it in the eqN of a line?

$2x-3y+7=0$ OR $2x-3y-7=0$ ?

Thanks for the help.
-NZF
July 30th 2006, 04:40 PM
Quick

Quote:

Originally Posted by 905

Hey,

I have this question thats been bugging me for a few days now.

$3y-15=2x-8$

Whats the answer (below) if we put it in the eqN of a line?

$2x-3y+7=0$ or $2x-3y-7=0$ ?

Thanks for the help.
-NZF

are you asking the solution of the line (when y equals 0) or are you asking for us to put the equation in standard/y-intercept/point-slope form?
July 30th 2006, 04:44 PM
NineZeroFive

Quote:

Originally Posted by Quick

are you asking the solution of the line (when y equals 0) or are you asking for us to put the equation in standard/y-intercept/point-slope form?

Put the equation in standard form. :p

EDIT: Actually it's just part of a question, the only part I don't fully understand.

-NZF
July 30th 2006, 04:48 PM
Quick

Quote:

Originally Posted by NineZeroFive

Put the equation in standard form. :p

alright, standard form is in the form of $ax+by=c$ therefore we start with your equation...

$3y-15=2x-8$

add 15 to both sides $3y=2x-8+15$

subtract 2x from both sides: $3y-2x=7$

voila!

~ $Q\!u\!i\!c\!k$
July 30th 2006, 04:50 PM
topsquark

Quote:

Originally Posted by NineZeroFive

Hey,

I have this question thats been bugging me for a few days now.

$3y-15=2x-8$

Whats the answer if we put it in the eqN of a line?

$2x-3y+7=0$ OR $2x-3y-7=0$ ?

Thanks for the help.
-NZF

Algebraically it would have to be $2x-3y+7=0$ . Other standard forms:
$y = \frac{2}{3}x + \frac{7}{3}$ (Slope - Intercept form)

and

$(y - 3) = \frac{2}{3}(x - 1)$ (An example of point-slope form)

and

$\frac{x}{3} - \frac{y}{2} = - \frac{7}{6}$ (I forget what this one is called.)

-Dan
July 30th 2006, 04:53 PM
NineZeroFive

Quote:

Originally Posted by Quick

alright, standard form is in the form of $ax+by=c$ therefore we start with your equation...

$3y-15=2x-8$

add 15 to both sides $3y=2x-8+15$

subtract 2x from both sides: $3y-2x=7$

voila!

~ $Q\!u\!i\!c\!k$

$3y-2x-7=0$ ? Because it the textbook it says its $2x-3y+7=0$ ?

:confused:
July 30th 2006, 04:56 PM
Quick

Quote:

Originally Posted by NineZeroFive

$3y-2x-7=0$ ? Because it the textbook it says its $2x-3y+7=0$ ?

:confused:

multiply both sides of $3y-2x-7=0$ by negative 1
July 30th 2006, 04:57 PM
NineZeroFive

Sorry,

I think I made a typo somewhere.

3y-15=2x-8 is the equation you need to put in the standard form.

Is the answer 2x-3y+7=0? (textbook answer) Why? -.-

Why is it NOT 2x-3y-7=0?

-NZF
July 30th 2006, 04:57 PM
topsquark

Quote:

Originally Posted by NineZeroFive

$3y - 15 = 2x - 8$

$3y - 15 = 2x - 8$

$3y - 15 - 3y = 2x - 8 - 3y$

$-15 = 2x - 3y - 8$

$-15 + 15 = 2x - 3y - 8 + 15$

$0 = 2x - 3y + 7$

-Dan
July 30th 2006, 05:01 PM
Quick

Quote:

Originally Posted by NineZeroFive

Sorry,

I think I made a typo somewhere.

3y-15=2x-8 is the equation you need to put in the standard form.

Is the answer 2x-3y+7=0? (textbook answer) Why? -.-

Why is it NOT 2x-3y-7=0?

-NZF

I assume you mean:

Quote:

Originally Posted by NineZeroFive

Sorry,

I think I made a typo somewhere.

3y-15=2x-8 is the equation you need to put in the standard form.

Is the answer 2x-3y+7=0? (textbook answer) Why? -.-

Why is it NOT 3y-2x-7=0?

-NZF

but the answer, both equations are correct, your book is just picky.
although I must say, I've never actually seen standard form equalling zero...
July 30th 2006, 05:08 PM
topsquark

Quote:

Originally Posted by Quick

I assume you mean:

but the answer, both equations are correct, your book is just picky.
although I must say, I've never actually seen standard form equalling zero...

One standard way of writing a multinomial (the "most" standard I've seen, if there is such a thing) is to write the terms of highest degree first down to the lowest degree such that the expression equals zero. For example:
$3x^2y^3 - xy^2 + 12x - 3y + 5 = 0$

In the case for a linear equation the form simply becomes:
$ax + by + c = 0$

I'll admit I don't usually see this form for a line, but considering the expression as a multinomial it would be standard.

-Dan
July 30th 2006, 05:09 PM
NineZeroFive

Quote:

Originally Posted by Quick

I assume you mean:

but the answer, both equations are correct, your book is just picky.
although I must say, I've never actually seen standard form equalling zero...

I'm still kinda confused because the book says "The equation of the line through A and B is 2x-3y+7=0. :confused:

Why is it 2x-3y+7=0 and NOT 2x-3y-7=0?

hmm..

-NZF
July 30th 2006, 05:11 PM
topsquark

Quote:

Originally Posted by NineZeroFive

I'm still kinda confused because the book says "The equation of the line through A and B is 2x-3y+7=0. :confused:

Why is it 2x-3y+7=0 and NOT 2x-3y-7=0?

hmm..

-NZF

Unless you have a question about my derivation, please see post #9 under this thread. I derived the form there.

-Dan