Question

There are m points on a straight line $AB$ and n points on another straight line $AC$ in which $A$ is not included. By joining these points triangles are constructed.
i) When $A$ is not included
ii) When $A$ is included

The ratio of number of triangles in both cases is

• A. $\frac{m+n-2}{m+n}$
• B. $\frac{m+n-2}{2}$
• C. $\frac{m+n-2}{m+n+2}$
• D. $\frac{m+n+2}{m+n-2}$

Answer 1

Answer: (A)

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

When $A$ is not in included then number of triangle formed by taking $2$ points on $AB$ and $1$ poins on $AC$ or $1$ point on $AB$ and $2$ points on $AC$.

${=}^m{C_2\times }^n{C_1+}^m{C_1\times }^n{C}_2=\frac{m(m-1).n}{2!}+\frac{mn(n-1)}{2!}$

$=\frac{1}{2}mn[m+n-2]$ ...(1)

When $A$ is included ,then except above triangles, there are triangles which have a point coincident with $A$ and $1$ vetex on $AB$ and $1$ vertex on $AC$.

$\therefore$ number of triangles of these type
${=}^m{C_1\times }^n{C}_1=mn$

when $A$ is included ,number of triangles

$=mn+\frac{mn}{2}[m+n-2]=\frac{1}{2}mn[m+n]$ ...(2)

The ratio of (1) and (2) $=\frac{m+n-2}{m+n}$