Question

How many lines can be drawn through three collinear points and three non- collinear points? Mark six points A,B,C,D,E and F such that A,B,C are collinear and D,E,F are non collinear.


full explanation please​

Answer 1

Step-by-step explanation:

For drawing a line we need at least two points.

If there are more than two points given we can still draw a line through all of them if there are all collinear.

And you said whichever line I have to draw should not pass through any three points. Or maybe in other words whichever line I choose to draw will not pass through any three points because no three points are collinear.

Of course in a plane if you choose any two points they have to be collinear, there is no other way out. You will not find two points in a plane which are not collinear.

So if you desire to draw any line passing through two points out of those 6 points, then just select any two points out of the six points. You will get a line out of that.

So basically the number of lines that can be drawn is simply equal to the number of ways in which we can choose two points out of those 6 points.

And if you know even a little bit of combinatorics, you may have knowledge that if you want to choose r things out of n things, then that can be done in nCr ways

So for question we need to choose two points out of 6. So that can be done in 6C2 ways.

Which is equal to 6!/(6–2)!(2!)=6x5/2x1=

15ways