Question

Six points in a plane be joining in all possible way by staright lines, and if no two of them be coincident or parallel, and no three pass through the same point (with the exception of the original $6$ points). The number of distinct points of intersection is equal to

• A. $105$
• B. $65$
• C. $51$
• D. $45$

Answer 1

The correct option is C $51$

Number of lines from $6$ points $={}^6{C}_2=15$.
Points of intersection obtained from these lines $={}^{15}{C}_2=105.$
Now we find the number of times, the original $6$ points come.
Consider one point say $A_1.$
Joining $A_1$ to remaining $5$ points, we get $5$ lines, and any two lines from these $5$ lines give $A_1$ as the point of intersection.
$\therefore A_1$ come ${}^5{C}_2=10$ times in $105$ points of intersection.
Similar is the case with other five points.
$\therefore 6$ original points comes $6\times 10=60$ times in points of intersection. Hence the number of distinct points of intersection $=105-60+6=51$