Question

There are $10$ points in a plane $4$ points are collinear. Other than the $4$ points, no other set of $3$ points is collinear. All points are joined to one another. Let $L$ be the number of different straight lines and $T$ be the number of different triangles, then

• A. $T=120$
• B. $L=40$
• C. $T=3L-4$
• D. $T=116$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $T=3L-4$
C $L=40$
D $T=116$

Out of $10$ points any 2 points can be selected in ${}^{10}{C}_2$ ways.

Out of $4$ collinear points, any 2 can be selected in ${}^4{C}_2$ ways.

Joining any 2 points gives a straight line. But for collinear points one line will be counted ${}^4{C}_2$ times.

Hence, total number of unique straight lines $L$ is ${}^{10}{C}_2-{(}^4{C}_2)+1=40$

Similarly, out of $10$ points any 3 points can be selected in ${}^{10}{C}_3$ ways.

Out of $4$ collinear points, any 3 can be selected in ${}^4{C}_3$ ways.

Since joining collinear points will not form a triangle, the number of different triangles $T$ is ${}^{10}{C}_3{-}^4{C}_3=116$

$3L-4=116=T$