# [SOLVED] Find the coordinates of three points on each of the following line.

• Mar 11th 2009, 10:08 AM
TJ2988
[SOLVED] Find the coordinates of three points on each of the following line.
Hi,
I just want a clarification of the question. Given the following symmetric equation:

http://www.virtualhighschool.com/con...uations/24.gif

do I have to derive to parametric equations so that I can sub in three different values for x,y, and z to get the three coordinates?

Thanks for any help, and sorry if it is the wrong place to post.
T.
• Mar 11th 2009, 10:44 AM
HallsofIvy
Quote:

Originally Posted by TJ2988
Hi,
I just want a clarification of the question. Given the following symmetric equation:

http://www.virtualhighschool.com/con...uations/24.gif

do I have to derive to parametric equations so that I can sub in three different values for x,y, and z to get the three coordinates?

Thanks for any help, and sorry if it is the wrong place to post.
T.

Apparently you have be registered with "virtualhighschool" to be able to view that site- and I'm not. But if it is just a symmetric equation, you can, but don't have to, derive parametric equations but if your equations are of the form f(x,y,z)= g(x,y,z)= h(x,y,z), you can separate that into two equations, f(x,y,z)= g(x,y,z) and g(x,y,z)= h(x,y,z). Choose whatever value you like for one of the variables, say x, put that into the two equations, and solve the two equations for y and z.

For example if you equations is 2x+ y- z= x+ y+ 2z= x- y+ z, you can separate them as 2x+ y- z= x+ 2y+ 2z and x+ y+ 2z= x- y+ z. Now:

If x= 0, these give y-z= 2y+ 2z and 2y+ 2z= y+ z. Those are equivalent to y+ 3z= 0 and y+ z= 0. The only solution to that is (0, 0, 0).

If x= 1, 2+ y- z= 1+ 2y+ 2z and 1+y+ 2z= 1- y+ z. Those are equivalent to y+ 3z= 1 and 2y+z= 0. z= -2y so y+ 3(-2y)= -5y= 1. y= -1/5 and z= 2/5. The point is (1, -1/5, 2/5).
• Mar 11th 2009, 12:25 PM
TJ2988
Oops, didn't realize the equation didn't show up.

The equation given is:

(x+2)/2 = (y-1)/2 = (z+5)/-3

So according to your example, I would separate the equations so that they read ... (x+2)/2 = (y-1)/2 and (y-1)/2 = (z+5)/-3

Although the equations do not follow the same layout as yours do.

I guess I can't take the parametric equations of:
x=-2+3t
y=1+2t
z=-5-3t
and choose values for t and find coordinates?

Thanks for the help so far.
• Mar 11th 2009, 01:00 PM
running-gag
Hi

Parametrization of the equations is a good idea
(x+2)/2 = (y-1)/2 = (z+5)/-3 = t

gives
x=2t-2
y=2t+1
z=-3t-5

Then you can chose any value for t
• Mar 11th 2009, 01:02 PM
TJ2988
thanks for the quick reply and help.
i just wrote down the x wrong, as its (x+2)/3 not 2, which is where I made a mistake making you make mistake (Giggle) ... haha
• Mar 11th 2009, 01:14 PM
running-gag
OK perfect (Wink)