# Equation of a line that passes through a point.

• May 21st 2011, 09:27 PM
Pupil
Equation of a line that passes through a point.
Find the equation of the line that passes through A(1,-4, 2) and is parallel to the intersection line of the two planes x - 2y + 3z - 1 = 0 and x - 4y+ 2z - 8 = 0.

What I did:
First I set the first and second equations to [1] and [2]:
x - 2y + 3z - 1 = 0 [1]
x - 4y+ 2z - 8 = 0 [2]
I then multiply [1] by 2 and use elimination to get rid of the y variable for now:
2x - 4y + 6z - 2 = 0
x - 4y + 2z - 8 = 0
________________________
x + 4z + 6 = 0 [3]

I'll then let z = t to solve for x in equation [3]:
x + 4t + 6 = 0
x = -4t - 6
Now I substitute z = t and x = -4t - 6 into equation [1] to solve for y:
-4t - 6 - 2y + 3t - 1 = 0
y = (-1/2)t - 7/2

Now that I have the values of all the unknowns, I first express it in parametric form:
x = -4t - 6
y = (-1/2)t - 7/2
z = t
Knowing this, finally, the direction vector for the line that passes through A(1, -4, 2) can be expressed:
(x,y,z) = (1, -4, 2) + t(-4, 1/2, 1)

I just wanted to know, did I do this correctly? I feel as if I did something wrong. If I did, can you point where I went wrong? Thank you in advance.
• May 22nd 2011, 01:59 AM
HallsofIvy
Looks good to me!

(Since the "direction vector", <-4, 1/2, 1>, shows only direction, its length is not important and any multiple of it would also work. If you really don't like fractions multiply it by 2: (x,y,z) = (1, -4, 2) + t(-8, 1, 2). That gives the same points, with this new t half what the old t was.)
• May 22nd 2011, 11:48 AM
Pupil
Quote:

Originally Posted by HallsofIvy
Looks good to me!

(Since the "direction vector", <-4, 1/2, 1>, shows only direction, its length is not important and any multiple of it would also work. If you really don't like fractions multiply it by 2: (x,y,z) = (1, -4, 2) + t(-8, 1, 2). That gives the same points, with this new t half what the old t was.)

Alrighty. Thanks HallsofIvy for all the help!