For every point P, there exists at least two distinct lines through P.
Please help me
Thanks
Mandy, Do you realize that every course in axiomatic geometry has a slightly different set of axioms? We do not know what yours are.
That said, you should have axioms somewhat like these.
Axiom: There are three non-collinear points.
Axiom: Given two points there is a line containing the points.
So you could say something like this.
Given a point P, by the first axiom there is another point Q.
By the second axiom there is a line $\displaystyle \ell _1 \;\& \;\left\{ {P,Q} \right\} \subseteq \ell _1 $.
Again by the first axiom $\displaystyle \left( {\exists R} \right)\left[ {R \notin \ell _1 } \right]$. Why is that true?
So $\displaystyle P\ne R$. Why is that true?
Can you find a second line $\displaystyle \ell _2 \;,\;\left\{ {P,R} \right\} \subseteq \ell _2 \;\& \;\ell _1 \ne \ell _2 $.
Sorry about that, my axioms are as follows
Incidence Axiom 1: For every point P and for every point Q not equal to P, there exists a unique line l incident with P and Q
Incidence Axiom 2: For every line l, there exist at least two distinct points incident with l
Incidence Axiom 3: There exist three distinct points with the property that no line is incident with all three of them.
So would the proof be as follows??
Given a point P, by IA3 there is another point Q.
By the IA2 there is a line .
Again by IA3 because no line is incident with all three of them
So because IA1 says there must be a unique line and if P=R there is not two unique lines where P is not equal to R
To find the second line we would go through what we did above, only substituting R in for Q and vice versa right?
Thanks for your time