# Simple eqN of a line question

• Jul 30th 2006, 03:35 PM
NineZeroFive
Simple eqN of a line question
Hey,

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

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

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

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

Thanks for the help.
-NZF
• Jul 30th 2006, 03:40 PM
Quick
Quote:

Originally Posted by 905
Hey,

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

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

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

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

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?
• Jul 30th 2006, 03:44 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
• Jul 30th 2006, 03:48 PM
Quick
Quote:

Originally Posted by NineZeroFive
Put the equation in standard form. :p

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

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

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

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

voila!

~ $\displaystyle Q\!u\!i\!c\!k$
• Jul 30th 2006, 03:50 PM
topsquark
Quote:

Originally Posted by NineZeroFive
Hey,

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

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

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

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

Thanks for the help.
-NZF

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

and

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

and

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

-Dan
• Jul 30th 2006, 03:53 PM
NineZeroFive
Quote:

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

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

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

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

voila!

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

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

:confused:
• Jul 30th 2006, 03:56 PM
Quick
Quote:

Originally Posted by NineZeroFive
$\displaystyle 3y-2x-7=0$? Because it the textbook it says its $\displaystyle 2x-3y+7=0$?

:confused:

multiply both sides of $\displaystyle 3y-2x-7=0$ by negative 1
• Jul 30th 2006, 03:57 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.

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

-NZF
• Jul 30th 2006, 03:57 PM
topsquark
Quote:

Originally Posted by NineZeroFive
$\displaystyle 3y - 15 = 2x - 8$

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

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

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

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

$\displaystyle 0 = 2x - 3y + 7$

-Dan
• Jul 30th 2006, 04:01 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.

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.

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...
• Jul 30th 2006, 04:08 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:
$\displaystyle 3x^2y^3 - xy^2 + 12x - 3y + 5 = 0$

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

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
• Jul 30th 2006, 04:09 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
• Jul 30th 2006, 04:11 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