Question

There are $m$ points on a straight line $AB$ and $n$ points on the line $AC$ none of them being the point $A$ . Triangles are formed with these points as vertices, when:
(i). $A$ is excluded
(ii).$A$ is included. The ratio of the number of triangles in the two cases is:
A.$\dfrac{{m + n - 2}}{{m + n}}$
B.$\dfrac{{m + n - 1}}{{m - n}}$
C.$\dfrac{{m + n - 2}}{{m + n + 2}}$
D.$\dfrac{{m\left( {n - 1} \right)}}{{\left( {m + 1} \right)\left( {n + 1} \right)}}$

Answer 1

For the first part if point A is excluded we can either choose 2 points on the line $AB$ (other than A) and 1 point on the line $AC$ (other than A) then the total number of possible triangles in this case $= \dfrac{{m\left( {m - 1} \right)}}{2} \times n$ and if we choose 2 points on the line $AC$ (other than A) and 1 point on the line $AB$ (other than A) then the total number of possible triangles in this case $= m \times \dfrac{{n\left( {n - 1} \right)}}{2}$ . Therefore the total number of triangles $= \dfrac{{mn}}{2}\left( {m + n - 2} \right)$ . For the second part, one point lies on line $AB$ and one on the straight line $AC$ , so the possible number of triangles is $mn$ . Then the ratio is $= \dfrac{{m + n - 2}}{2}$

Complete answer:
(i).In the problem, we are given that there are $m$ points on the straight line $AB$ and $n$ points on the straight line $AC$ .
To form a triangle we need to choose three non-collinear points on the two straight lines $AB$ and $AC$ .
So, if point A is excluded we can either choose 2 points on the line $AB$ (other than A) and 1 point on the line $AC$ (other than A).
In this case the total number of possible triangles in this case $= \dfrac{{m\left( {m - 1} \right)}}{2} \times n$
Or
We can either choose 2 points on the line $AC$ (other than A) and 1 point on the line $AB$ (other than A)
In this case the total number of possible triangles in this case $= m \times \dfrac{{n\left( {n - 1} \right)}}{2}$
So, the total number of possible triangles $= \dfrac{{m\left( {m - 1} \right)}}{2} \times n + m \times \dfrac{{n\left( {n - 1} \right)}}{2}$
$= \dfrac{{mn}}{2}\left( {m - 1 + n - 1} \right)$
$= \dfrac{{mn}}{2}\left( {m + n - 2} \right)$
(ii).If point A is also included then choose two non-collinear points, one point on a straight line $AB$ and one point on a straight line $AC$ .
So, the total number of triangles that can be formed $= mn$
Therefore, the ratio of the possible number of triangles that can be formed excluding point A and the possible number of triangles that can be formed including point A $= \dfrac{{\dfrac{{mn}}{2}\left( {m + n - 2} \right)}}{{mn}}$
$= \dfrac{{m + n - 2}}{2}$

Note:
In the solution to reach the answer we considered every triangle that could be possibly made with the arrangement given to us and then choose what number of vertices could be used to make those triangles, that is how we reached our solution in both the parts. To solve such problems first consider the number of points on one line and then multiply it with the possible number of points on the other line.