SSC Sample Paper Mock Test-7 SSC CGL Tear-II Paper-1

Find the ratio in which the point P\[\left( \frac{3}{4},\frac{5}{12} \right)\] the line segment joining the points A\[\left( \frac{1}{2},\frac{3}{2} \right)\]

A) 1 : 5

B) 5 : 1

C) 3 : 2

D) 4 : 5

Correct Answer: A

Solution :

Let P \[\left( \frac{3}{4},\frac{5}{12} \right)\] divide AB internally in the ratio m: n. Using the section formula, we get

\[\left( \frac{3}{4},\frac{5}{12} \right)=\left( \frac{2\,m-\frac{n}{2}}{m+n},\frac{-5\,m+\frac{3}{2}n}{m+n} \right)\]

On equating, we get

\[\Rightarrow \]\[\frac{3}{4}=\frac{2\,m-\frac{n}{2}}{m+n}\]

and\[\frac{5}{12}=\frac{-\,\,5\,m+\frac{3}{2}n}{m+n}\]

\[\Rightarrow \]\[\frac{3}{4}=\frac{4\,m-n}{2\,\,(m+n)}\]

and\[\frac{5}{12}=\frac{-\,\,10\,m+3\,\,n}{2\,\,(m+n)}\]

\[\Rightarrow \] \[\frac{3}{2}=\frac{4\,m-n}{m+n}\]and \[\frac{5}{6}=\frac{-\,\,10\,m+3\,n}{m+n}\]

\[\Rightarrow \] \[x=1\]

\[\Rightarrow \] \[5\,m+5\,n=-60\,m+18\,n\]

\[\Rightarrow \] \[5\,n-5\,m=0\]and \[65\,\,m-13\,\,n=0\]

\[\Rightarrow \] \[n=m\]and\[13\,\,(5\,m-n)=0\]

\[\Rightarrow \] \[n=m\]and \[5\,\,m-n=0\]

\[\because \] \[m=n\] does not satisfy.

\[\therefore \] \[5\,m-n=0\]

\[\Rightarrow \] \[5\,\,m=n\]

\[\Rightarrow \] \[\frac{m}{n}=\frac{1}{5}\]