There are m points on a straight line ab

Asked by admin @ 20/08/2021 in Math viewed by 320 People

There are m points on a straight line AB and n points on another line AC, none of them being the point A. Triangle are formed from these points as vertices when

1) Aexcluded
2) A in included. the ratio of numbers of triangles in the two cases is :
A) m+n-2/m+n
b) m+n-2/m+n-1
c)m+n-2/m+n+2
d)m(n-1)/(m+1)(n+1)
( iit level question so not going to be that easy )

Answered by admin @ 20/08/2021

Answer:

Step-by-step explanation:

Given that there are m points on a straight line AB

Given that there are n points on a straight line AC

To form a triangle we need to choose 3 non-collinear points

1) If point A is excluded:

to form a triangle ,

we need to choose 2 points on the line AB(other than A) and 1 point on the line AC other than A

we need to choose 2 points on the line AC and 1 point on the line AB other than A.

Choosing 2 points on the line AB(other than A) can be done in mC2 ways , and 3rd vertex should be chosen on lie AC(other than A) which could be any one from the n points, hence both could happen in mc2*n ways

= mn(m-1)/2

Choosing 2 points on the line AC(other than A) can be done in nC2 ways , and 3rd vertex should be chosen on lie AB(other than A) which could be any one from the m points, hence both could happen in nc2*m ways

= mn(n-1)/2

Thus, total possible number of triangles that could be formed when A is exluded are mn(m-1)/2 + mn(n-1)/2

=mn(m+n-2)/2-----(*)

2)If A is included in the list of given points, then 2 possibilities arise

(i) If point A is chosen, then the other 2 vertices should be chosen each on different lines to avoid collinearity, hence one point from line AB(which can be done in 'm ways and the other point on AC which could be done in 'n' ways, hence selection of all vertices could be done in m*n ways.

(ii)If point A is not selected, then the selection is same as the first case, so total number of ways would be mn(m+n-2)/2.

Hence, total number of possibilities if point A is included will be

mn(m+n-2)/2 + mn = mn(m+n)/2----(**).

Hence ratio would be [mn(m+n-2)/2]/[mn(m+n)]

=(m+n-2)/(m+n).

There are m points on a straight line ab

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