Why Did I Get the Wrong Parametric Equation for L1?

[roam]

Find the parametric equations of the lines L₁, L₂, L₃ and L₄ that pass through the indicated verticies of the square:

http://img516.imageshack.us/img516/4052/parametrickl6.jpg


This isn't a homework question but I need some tutoring help here with parametric equations of lines, those who are good at linear algebra. Thanks.




3. The Attempt at a Solution

L₁ for example, is parallel to the y-axis a vector that's (0,1)

A vector equation of the line is x=tv. If we let x = (x,y) this equation can be expressed in the component form as;

(x,y) = t(0,1)

(t is some scalar.)

Therefore the parametric equation has to be:

x = t.0
y = t

\Rightarrow x = 0, y = t

But the correct answer in the textbook suggests: L₁: x = 1, y = t

Why did I get the wrong answer for x?

I'm pretty confused right now I've been reading the book this afternoon and I still don't get it, I’m too tired to continue, I’m off to bed now!

 

[danago]

You are right that L₁ is parallel to the vector (0,1), but when t=0, your equation suggest that (x,y)=(0,0) which cannot be correct, since the line does not pass through the origin. Can you see the detail you have overlooked?

 

[Defennder]

A vector equation of the line is x=tv. If we let x = (x,y) this equation can be expressed in the component form as;

(x,y) = t(0,1)

(t is some scalar.)

Therefore the parametric equation has to be:

x = t.0
y = t

\Rightarrow x = 0, y = t

But the correct answer in the textbook suggests: L₁: x = 1, y = t

No it isn't. A vector equation is of the form \vec{OR} + t\vec{v}. Assuming that you meant that the x-component of the line is x=tv where v is a vector, it's still incorrect because x is constant for L₁, and doesn't vary for any parameter t.