Question

10 points on straight line on ab and 8points on straight line ac then how many number of triangles are formed

Answer 1

Answer:

504

Step-by-step explanation:

let Point - A is common on both lines .

total no of ∆ = ∆ of type-1 + ∆ of type-2 + ∆ of type-3

= 63 + 189 + 252 = 504

∆ of type-1 = A as one vertex and other two vertices on different lines

= 1 × 9C1 × 7C1

= 63

now, excluding point A :

∆ of type-2 = one vertex on AB and other two vertices on AC

= 9C1 × 7C2

= 9 × 7×6/2

= 189

∆ of type-3 = one vertex on AC and other two vertices on AB

= 7C1 × 9C2

= 7 × 9×8/2

= 252