Question

There are $10$ points in a plane, no three of which are in the same straight line excepting $4$, which are collinear. Then number of

• A. Straight lines formed by joining them is $40$
• B. Triangles formed by joining them is $116$
• C. Straight lines formed by joining them is $45$
• D. Triangle formed by joining them is $120$

Answer 1

Answer: (A), (B)

The correct options are
A Straight lines formed by joining them is $40$
B Triangles formed by joining them is $116$

We know that joining of any 2 points give a line. Thus the number of lines obtained from 10 points, when no 3 of which are collinear $={{}^{10}C}_2=45$

Lines obtained from 4 points $={{}^{4}C}_2=6$

No of lines lost due to 4 collinear points $=6-1=5$

So required number of lines $=45-5=40$

Also we know that any triangle can be obtained by joining any 3 points not in the same straight line. Thus number of triangles obtained from 10 different point, no 3 of which are collinear are $={{}^{10}C}_3=120$

Triangles obtained from 4 points $={{}^{4}C}_3=4$

Number of triangles lost due to 4 collinear points=4

So required number of triangles $=120-4=116$