There are m points on a straight line AB and | vedclass

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Question

(A) Case $I$: When $A$ is excluded.To form a triangle,we must select $3$ non-collinear points. Since $m$ points are on $AB$ and $n$ points are on $AC$

Vedclass · Accepted Answer

$\frac{m + n - 2}{m + n}$

Vedclass · Answer

(A) Case $I$: When $A$ is excluded.To form a triangle,we must select $3$ non-collinear points. Since $m$ points are on $AB$ and $n$ points are on $AC$,we can select $2$ points from $AB$ and $1$ from $AC$,or $1$ from $AB$ and $2$ from $AC$.Number of triangles $= ^mC_2 \cdot ^nC_1 + ^mC_1 \cdot ^nC_2 = \frac{m(m-1)}{2} \cdot n + m \cdot \frac{n(n-1)}{2} = \frac{mn}{2}(m - 1 + n - 1) = \frac{mn(m + n - 2)}{2}$.Case $II$: When $A$ is included.If $A$ is a vertex,we need to select $2$ more points. These can be $1$ from $AB$ and $1$ from $AC$,or $2$ from $AB$ (excluding $A$),or $2$ from $AC$ (excluding $A$).Number of triangles $= (m \cdot n) + ^mC_2 + ^nC_2 = mn + \frac{m(m-1)}{2} + \frac{n(n-1)}{2}$.Alternatively,total triangles including $A$ is the sum of triangles formed by $A$ and one point from each line $(mn)$ plus the triangles formed by $A$ and two points from one line $(^mC_2 + ^nC_2)$.Total $= \frac{mn(m+n-2)}{2} + mn + ^mC_2 + ^nC_2 = \frac{mn(m+n-2) + 2mn + m(m-1) + n(n-1)}{2} = \frac{mn(m+n) + m^2-m + n^2-n}{2}$.Wait,the standard interpretation for this problem is: Case $I$ is $\frac{mn(m+n-2)}{2}$. Case $II$ is the total triangles possible using $A$ as a vertex,which is $mn + ^mC_2 + ^nC_2$. However,based on the provided options,the ratio is $\frac{m+n-2}{m+n}$.

There are $m$ points on a straight line $AB$ and $n$ points on another line $AC$,none of them being the point $A$. Triangles are formed from these points as vertices when $(i)$ $A$ is excluded and $(ii)$ $A$ is included. Then the ratio of the number of triangles in the two cases is

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