Question

How many lines are determined by the points a b c d and e if 3 of them arent collinear

Answer 1

Answer:

10 possible lines

Step-by-step explanation:

Given

$Points = \{a,b,c,d,e\}$

Required

Number of lines

From the question, the number of points (n) are:

$n = 5$

If 3 points are not collinear (i.e. on a straight line), it means that only (5 - 3) of the points can be chosen

So, we have:

$n = 5$

$r = 5 - 3$

$r =2$

The number of lines is then calculated using combination formula.

$^nC_r = \frac{n!}{(n-r)!r!}$

We used combination because the analysis done above implies that 2 points are to be selected from a,b,c,d and e.

To select means combination

Having said that;

Substitute $5\ for\ n\ and\ 2\ for\ r$

$^5C_2 = \frac{5!}{(5-2)!2!}$

$^5C_2 = \frac{5!}{3!2!}$

$^5C_2 = \frac{5*4*3!}{3!2!}$

$^5C_2 = \frac{5*4}{2!}$

$^5C_2 = \frac{5*4}{2*1}$

$^5C_2 = \frac{20}{2}$

$^5C_2 = 10$

Hence, there are 10 possible lines