# parallel & perpendicular lines help

• Oct 17th 2008, 02:32 PM
shipwreck
parallel & perpendicular lines help
hey everyone, this is my first post...i hope i can find some help here, as i have searched my math book up and down, only to find no answer.

i am in college algebra, and the section is parallel & perpendicular lines.

the problem i am totally stuck on is this...

Code:

 x - y      x + y -----  =  -----  - 1   3            2   AND   7 = - ( x - y ) + 4y
They want me to rewrite both in slope-intercept form, then find out if they are parallel or perpendicular.

I would really and truly appreciate any and all help with this...i just don't know what to do

thanks so much,

ryan
• Oct 17th 2008, 02:48 PM
masters
Quote:

Originally Posted by shipwreck
hey everyone, this is my first post...i hope i can find some help here, as i have searched my math book up and down, only to find no answer.

i am in college algebra, and the section is parallel & perpendicular lines.

the problem i am totally stuck on is this...

Code:

 x - y      x + y -----  =  -----  - 1   3            2   AND   7 = - ( x - y ) + 4y
They want me to rewrite both in slope-intercept form, then find out if they are parallel or perpendicular.

I would really and truly appreciate any and all help with this...i just don't know what to do

thanks so much,

ryan

Hello Ryan,

$\frac{x-y}{3}=\frac{x+y}{2}-1$

First, let's multiply everything by the LCD of 6 to get rid of those pesky denominators.

$6\left(\frac{x-y}{3}\right)=6\left(\frac{x+y}{2}\right)-6(1)$

$2x-2y=3x+3y-6$

Simplify to

$-5y=x-6$

$\boxed{y=\frac{-1}{5}x+\frac{6}{5}}$

Now for the other one...

$7=-(x-y)+4y$

$7=-x+y+4y$

Simplifying....

$5y=x+7$

$\boxed{y=\frac{1}{5}x+\frac{7}{5}}$

Compare the slopes of the two boxed equations. What do you think?
• Oct 17th 2008, 02:53 PM
shipwreck
well i know that if both slopes are identical, than they are considered parallel....but these aren't identical, although i don't know if there is a particular rule or not about them being opposites, but NOT being reciprocal negative opposites...

1/5 x & 1/5x would be parallel, but i am just not 100% certain about 1/5x & -1/5x. i would say they are not parallel...right?
• Oct 17th 2008, 03:29 PM
shipwreck
anyone able to confirm if i am right or not about these being parallel or not?
• Oct 17th 2008, 03:34 PM
11rdc11
Quote:

Originally Posted by shipwreck
anyone able to confirm if i am right or not about these being parallel or not?

You answered your question yourself(Rofl). If the slopes are not the same then they aren't the parallel