Why Did I Get the Wrong Parametric Equation for L1?

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The discussion centers on the derivation of parametric equations for lines L1, L2, L3, and L4, specifically addressing the confusion surrounding line L1. The user initially proposed the parametric equation as x = 0, y = t, which incorrectly suggests that L1 passes through the origin. However, the correct parametric equation for L1 is x = 1, y = t, indicating that L1 is parallel to the y-axis and does not vary in the x-direction. The misunderstanding arises from the misapplication of vector equations in linear algebra.

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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

[tex]\Rightarrow x = 0, y = t[/tex]

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!
 
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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?
 
roam said:
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

[tex]\Rightarrow x = 0, y = t[/tex]

But the correct answer in the textbook suggests: L1: x = 1, y = t
No it isn't. A vector equation is of the form [tex]\vec{OR} + t\vec{v}[/tex]. 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.
 

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