# Thread: points proof

1. ## points proof

1) Let L be a line in coordinate 2- space R^2 or 3- space R^3, let x be a point not on L, and let p1,...pn be points on L. Prove that the lines xp1,...xpn are distinct. Why does this imply that R^2 and R^3 contain infinitely many lines?

2) Suppose that L1,...Ln are lines in coordinate 2- space R^2 or 3- space R^3. Prove that there is a point q which does not lie on any of these lines. (hint: Take a line M which is different from each of L1,...Ln; for each j we know that M and Lj have at most one point in common, but we also know how M has infinitely many points.)

Thanks a whole lot.

2. Hi ThePerfectHacker,

Thank you so much for trying to help me out but I don't think that is how I am suppose to prove it. Still, thanks a lot.

Jen

3. Originally Posted by jenjen
1) Let L be a line in coordinate 2- space R^2 or 3- space R^3, let x be a point not on L, and let p1,...pn be points on L. Prove that the lines xp1,...xpn are distinct. Why does this imply that R^2 and R^3 contain infinitely many lines?

2) Suppose that L1,...Ln are lines in coordinate 2- space R^2 or 3- space R^3. Prove that there is a point q which does not lie on any of these lines. (hint: Take a line M which is different from each of L1,...Ln; for each j we know that M and Lj have at most one point in common, but we also know how M has infinitely many points.)

Thanks a whole lot.
Since the questions concerns coordinate space, are you supposed to use vectors?

For question 1, you could treat a line through points $x$ and $p_n$ as a parametric equation $v_n (t) = (p_n - x)t + x.$ Then to show the lines are distinct, show $p_i \ne p_j$ implies $v_i (t_i) \ne v_j (t_j)$ except when $t_i = t_j = 0.$ To do this takes a little work, but basically you treat the line L as a parametric equation $w (t) = (p_i - p_j)t + p_j$ and then show $v_i (t_i) = v_j (t_j)$ would imply one of two cases. Case 1 is $t_i = t_j$ and $p_i = p_j .$ Case 2 is $x = w(t)$ for $t = t_i/(t_i-t_j)$ and thus $x$ is on the line L. Either case contradicts the assumptions.