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I can get up to showing that it is an abstract geometry, but I cannot figure out how to show that for every two points, there is a unique line. The definition of a line in vector form is given $L (AB) = \{X \in R \times R~ | ~X=A+t(B-A) ~\text{for some}~~ t \in \mathbb R\}$

$A,B,X$ are points.

I am trying to go about this by supposing A,B are in $L (CD)$ and $L (XY)$. From there I try to get $C=X$ and $D=Y$, but I can't do this. Any help or hint is appreciated.

The Cartesian Plane is the set of Points in $R \times R$ and the set of all lines $L (AB)$.

This is a problem in a textbook but I am not doing this for homework. I just find this subject really interesting so any help is appreciated.

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2 Answers 2

No, you can't get $C=X$ and $D=Y$. But you should be able to show that if $A,B\in L(CD)$ and $A,B\in L(XY)$, then $L(CD)\subset L(XY)$ (and so, symmetrically, $L(XY)\subset L(CD)$), whence the lines are equal. Here's a start: \begin{align*} A &= C + s(D-C) \quad\text{ for some } s\in\mathbb R\,; \\ B &= C + t(D-C) \quad\text{ for some } t\in\mathbb R\,; \\ A &= X + u(Y-X) \quad\text{ for some } u\in\mathbb R\,; \text{ and} \\ B &= X + v(Y-X) \quad\text{ for some } v\in\mathbb R\,. \end{align*} From this you should be able to show that $C\in L(XY)$ and likewise for $D$.

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You can't hope to get $C=X$ and $D=Y$, because any two points on a line determine that line. What you have to show is that $L(CD)=L(XY)$. In fact they are both $L(AB)$.

You have $$A=C+s_A(D-C)=X+t_A(Y-X)$$ and $$B=C+s_B(D-C)=X+t_B(Y-X)$$ for some $s_A,s_B,t_A,t_B\in\Bbb R$. Then $B-A=(s_B-s_A)(D-C)$, so

$$D-C=\frac 1{s_B-s_A}(B-A)\;,$$

and

$$C=A-s_A(D-C)=A-\frac{s_A}{s_B-s_A}(B-A)\;.$$

(Here I use the assumption that $A\ne B$.) In similar fashion you can express $D,X$, and $Y$ in terms of $A$ and $B$ to show that $C,D,X,Y\in L(AB)$. Once you have that, a little more work of the same kind will allow you to express every point of $L(CD)$ and $L(XY)$ as a point of $L(AB)$.

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Thank you for the help! I also appreciate that you didn't solve the problem, then I wouldn't have learned anything. –  Matthew Evers May 23 '13 at 19:26
    
@Matthew: You’re welcome. –  Brian M. Scott May 23 '13 at 20:14
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