Question

three noncollinear points determine three lines, as shown in the figure
(a) how many lines are determined by four points, no three of which are collinear?
(b) how many lines are determined by five points, no three of which are collinear?
(c) how many lines are determined by n points? assume that no three points are collinear

(a) four points determine \\(\square\\) lines

Explanation:

Step1: Understand line-point relationship

A line is determined by 2 distinct points, and we need to count unique pairs of points (since no three are collinear, no two pairs form the same line). This is a combination problem, calculated by the combination formula $C(n,k)=\frac{n!}{k!(n-k)!}$, where $n$ is total points, $k=2$ (points per line).

Step2: Solve for 4 points

Substitute $n=4, k=2$ into the formula:
$C(4,2)=\frac{4!}{2!(4-2)!}=\frac{4\times3\times2!}{2!\times2\times1}=\frac{4\times3}{2\times1}=6$

Step3: Solve for 5 points

Substitute $n=5, k=2$ into the formula:
$C(5,2)=\frac{5!}{2!(5-2)!}=\frac{5\times4\times3!}{2!\times3!}=\frac{5\times4}{2\times1}=10$

Step4: Solve for n points

Generalize with $n$ points, $k=2$:
$C(n,2)=\frac{n!}{2!(n-2)!}=\frac{n(n-1)(n-2)!}{2\times1\times(n-2)!}=\frac{n(n-1)}{2}$

Answer:

(a) 6
(b) 10
(c) $\frac{n(n-1)}{2}$