I am submitting this for the second time, because the person that answered the first time...

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Basic Math

I am submitting this for the second time, because the personthat answered the first time definitely did not read my last twoparagraphs. I need to DISCUSS, why the theorem works when adding apoint D. I cannot draw a triangle and throw point D on it. Iunderstand up to a certain point, but I have no idea what myprofessor is looking for in my answer. I have included her commentsto hopefully help you help me. Thank you.

My question is based on the following:

Consider the axiomatic system and theorem below:

Axiom 1: If there is a pair of points, then they are on a linetogether.

Axiom 2: If there is a line, then there must be at least twopoints on it.

Axiom 3: There exist at least three distinct points.

Axiom 4: If there is a line, then not all of the points can beon it,

Theorem 1: Each point is on at least two distinct lines.

I have proven and understand up to 3 points, but I am strugglingwith explaining what happens with the 4th point.

If I use Axiom 3 to create 4th point D (the first 3 being A, B,and C), this will give me distinct lines AD, BD, CD, ADB, ADC, andBDC.

ADB, ADC, and BDC were all previously existing lines, and AD,BD, and CD are new lines, correct?

These are two separate cases because I cannot have point D on apreviously existing line, and a new line, at the same time. I feelI understand up to this point. I need to discuss these twopossibilities separately, but I am confused on how to go aboutthat.

My professor states that I need to "discuss the differentpossibilities for how many distinct lines those are, we do not knowif those are 3 distinct lines or not, this is where the differentcases come in". I don't understand AT ALL what she is lookingfor.

Answer & Explanation Solved by verified expert
3.7 Ratings (455 Votes)
Please think about a square or rectangle Explanations According to Axiom 1 these pair of points are on the edges of a line together and these points are connected with two another sides of the    See Answer
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