Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   Introductory Physics Homework (http://www.physicsforums.com/forumdisplay.php?f=153)
-   -   Normal and Parallel Lines (http://www.physicsforums.com/showthread.php?t=56516)

ashleyk Dec12-04 09:12 PM

Normal and Parallel Lines
 
Find an equation of each line normal to the graph y=2x/(x-1) and parallel to the line 2x-y+1=0

HallsofIvy Dec12-04 09:44 PM

Why?

Could you give us some indication as to what you DO understand about these problems and what you have already tried yourself?

In particular, are these two separate problems or do you mean to find lines that are both normal to y= 2x/(x-1) AND parallel to 2x- y+ 1= 0?

Pyrrhus Dec12-04 10:25 PM

Use the concept of slope to solve this.

ashleyk Dec13-04 05:00 AM

I understand that a normal line is perpendicular and I know what the parallel line is. I know that you have to take a derivative to get the slope. But I'm assumimg the equation must be both normal and parallel(?)

Physicsisfun2005 Dec13-04 05:57 AM

hmmm maybe by parallel they are refering to a tangent line???......well.......maybe lol :smile:

HallsofIvy Dec13-04 06:31 AM

Quote:

Quote by ashleyk
I understand that a normal line is perpendicular and I know what the parallel line is. I know that you have to take a derivative to get the slope. But I'm assumimg the equation must be both normal and parallel(?)


Good! Now DO it! 2x-y+1= 0 is the same as y= 2x+ 1. What is the slope of that line? What is the slope of any line parallel to that?

Find the derivative of y= 2x/(x-1) (as a function of x- you don't yet know what x is). Calling that m(x), the slope of the normal line is -1/m(x). Set that equal to the slope you got above and solve for x to find the point(s) at which the normal is parallel to 2x-y+1= 0.

ehild Dec13-04 07:20 AM

Quote:

Quote by ashleyk
Find an equation of each line normal to the graph y=2x/(x-1) and parallel to the line 2x-y+1=0


Is it meant as shown in the attachment?

ehild


All times are GMT -5. The time now is 04:50 AM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums