Question 476372
You are making this question way too complicated. If your textbook is asking all those questions then I will presume that the textbook is making this question too complicated.


I think I have already solved this problem a few days ago, but I will solve it again for clarification. We form lines by choosing pairs of points (since two points uniquely determine a line). We have three cases:


Case 1: We choose two of the four collinear points. Here, only one line is determined because we cannot count 4C2 lines (they are all the same).


Case 2: We choose two of the six other non-collinear points. Here, it is simply choosing any two points out of 6, 6C2 = 15.


Case 3: We choose one of the four collinear points, and one of the six remaining points. There are 4*6 = 24 lines determined, no two of which are collinear.


The number of lines is 1+15+24 = 40.

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Okay, now for your "questions." #1 - yes. #2 - yes, we choose two points from four, 4C2 = 6. #3 and #4 are essentially the same. You subtract 4C2 = 6 because you are counting the same line six times, when you only want to count it once (hence, add 1 afterwards). #5 - yes, just did that. #6 - This means that you have four points that lie on the same line, and the other six are completely random and do not lie on this line. #7 - Yes, correct? #8 - "Collinear" means "on the same line."