Notes On Section Formula - CBSE Class 12 Maths
The position vector for R in both these cases. Case I: R divides AB internally.   $\stackrel{â†’}{\text{OA}}$  =            = $\stackrel{â†’}{\text{b}}$       = $\stackrel{â†’}{\text{r}}$    Point R divide line segment AB in the ratio m:n AR:RB = m:n In âˆ†OAR: $\stackrel{â†’}{\text{OR}}$   = $\stackrel{â†’}{\text{OA}}$   +  $\stackrel{â†’}{\text{AR}}$         â‡’ $\stackrel{â†’}{\text{AR}}$ = $\stackrel{â†’}{\text{OR}}$ - $\stackrel{â†’}{\text{OA}}$ â‡’ $\stackrel{â†’}{\text{AR}}$ = $\stackrel{â†’}{\text{r}}$ - $\stackrel{â†’}{\text{a}}$ In âˆ†ORB: $\stackrel{â†’}{\text{OB}}$ = $\stackrel{â†’}{\text{OR}}$ + $\stackrel{â†’}{\text{RB}}$ â‡’ $\stackrel{â†’}{\text{RB}}$ = $\stackrel{â†’}{\text{OB}}$ - $\stackrel{â†’}{\text{OR}}$ â‡’ $\stackrel{â†’}{\text{RB}}$ = $\stackrel{â†’}{\text{b}}$ - $\stackrel{â†’}{\text{r}}$ $\stackrel{â†’}{\text{AR}}$ :$\stackrel{â†’}{\text{RB}}$ = m:n â‡’ m $\stackrel{â†’}{\text{RB}}$ = n $\stackrel{â†’}{\text{AR}}$ â‡’ m ($\stackrel{â†’}{\text{b}}$ - $\stackrel{â†’}{\text{r}}$) = n ( $\stackrel{â†’}{\text{r}}$   - $\stackrel{â†’}{\text{a}}$  ) â‡’ m $\stackrel{â†’}{\text{b}}$ - m $\stackrel{â†’}{\text{r}}$ = n $\stackrel{â†’}{\text{r}}$ - n $\stackrel{â†’}{\text{a}}$ â‡’ m $\stackrel{â†’}{\text{r}}$ + n $\stackrel{â†’}{\text{r}}$= m $\stackrel{â†’}{\text{b}}$ + n $\stackrel{â†’}{\text{a}}$ â‡’ (m + n )$\stackrel{â†’}{\text{r}}$= m $\stackrel{â†’}{\text{b}}$ + n $\stackrel{â†’}{\text{a}}$ â‡’ $\stackrel{â†’}{\text{r}}$=  $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{+ n}\stackrel{â†’}{\text{a}}}{\text{m+n}}$  â‡’ $\stackrel{â†’}{\text{OR}}$ = $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{+ n}\stackrel{â†’}{\text{a}}}{\text{m+n}}$  The point R lies at the midpoint of line segment AB. When m : n = 1:1 â‡’ m = n â‡’ $\stackrel{â†’}{\text{OR}}$ =   $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{+ n}\stackrel{â†’}{\text{a}}}{\text{m+n}}$  =      â‡’ $\stackrel{â†’}{\text{OR}}$ =      $\frac{\text{m (}\stackrel{â†’}{\text{a}}\text{+}\stackrel{â†’}{\text{b}}\text{)}}{\text{2m}}$          â‡’ $\stackrel{â†’}{\text{OR}}$ =    =        Case II: R divides AB externally   $\stackrel{â†’}{\text{OA}}$  =            = $\stackrel{â†’}{\text{b}}$       = $\stackrel{â†’}{\text{r}}$    Let R divide AB externally such that AR:BR = m:n In âˆ†OAR: $\stackrel{â†’}{\text{OR}}$  = $\stackrel{â†’}{\text{OA}}$  + $\stackrel{â†’}{\text{AR}}$  â‡’ $\stackrel{â†’}{\text{AR}}$  = $\stackrel{â†’}{\text{OR}}$  - $\stackrel{â†’}{\text{OA}}$  â‡’ $\stackrel{â†’}{\text{AR}}$  = $\stackrel{â†’}{\text{r}}$  - $\stackrel{â†’}{\text{a}}$  In âˆ†OBR: $\stackrel{â†’}{\text{OR}}$  = $\stackrel{â†’}{\text{OB}}$  + $\stackrel{â†’}{\text{BR}}$  â‡’ $\stackrel{â†’}{\text{BR}}$  = $\stackrel{â†’}{\text{OR}}$  - $\stackrel{â†’}{\text{OB}}$  â‡’ $\stackrel{â†’}{\text{BR}}$  = $\stackrel{â†’}{\text{r}}$  - $\stackrel{â†’}{\text{b}}$  $\stackrel{â†’}{\text{AR}}$  :$\stackrel{â†’}{\text{BR}}$  = m:n â‡’ m ($\stackrel{â†’}{\text{BR}}$ )= n ($\stackrel{â†’}{\text{AR}}$ ) â‡’ m ($\stackrel{â†’}{\text{r}}$  - $\stackrel{â†’}{\text{b}}$ ) = n ($\stackrel{â†’}{\text{r}}$  - $\stackrel{â†’}{\text{a}}$  ) â‡’ m $\stackrel{â†’}{\text{r}}$ - m $\stackrel{â†’}{\text{b}}$ = n $\stackrel{â†’}{\text{r}}$ - n $\stackrel{â†’}{\text{a}}$ â‡’ m $\stackrel{â†’}{\text{r}}$ - n $\stackrel{â†’}{\text{r}}$= m $\stackrel{â†’}{\text{b}}$ - n $\stackrel{â†’}{\text{a}}$ â‡’ (m - n )$\stackrel{â†’}{\text{r}}$ = m $\stackrel{â†’}{\text{b}}$ - n $\stackrel{â†’}{\text{a}}$ â‡’ $\stackrel{â†’}{\text{r}}$ = $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{- n}\stackrel{â†’}{\text{a}}}{\text{m-n}}$  â‡’ $\stackrel{â†’}{\text{OR}}$ = $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{- n}\stackrel{â†’}{\text{a}}}{\text{m-n}}$ .

#### Summary

The position vector for R in both these cases. Case I: R divides AB internally.   $\stackrel{â†’}{\text{OA}}$  =            = $\stackrel{â†’}{\text{b}}$       = $\stackrel{â†’}{\text{r}}$    Point R divide line segment AB in the ratio m:n AR:RB = m:n In âˆ†OAR: $\stackrel{â†’}{\text{OR}}$   = $\stackrel{â†’}{\text{OA}}$   +  $\stackrel{â†’}{\text{AR}}$         â‡’ $\stackrel{â†’}{\text{AR}}$ = $\stackrel{â†’}{\text{OR}}$ - $\stackrel{â†’}{\text{OA}}$ â‡’ $\stackrel{â†’}{\text{AR}}$ = $\stackrel{â†’}{\text{r}}$ - $\stackrel{â†’}{\text{a}}$ In âˆ†ORB: $\stackrel{â†’}{\text{OB}}$ = $\stackrel{â†’}{\text{OR}}$ + $\stackrel{â†’}{\text{RB}}$ â‡’ $\stackrel{â†’}{\text{RB}}$ = $\stackrel{â†’}{\text{OB}}$ - $\stackrel{â†’}{\text{OR}}$ â‡’ $\stackrel{â†’}{\text{RB}}$ = $\stackrel{â†’}{\text{b}}$ - $\stackrel{â†’}{\text{r}}$ $\stackrel{â†’}{\text{AR}}$ :$\stackrel{â†’}{\text{RB}}$ = m:n â‡’ m $\stackrel{â†’}{\text{RB}}$ = n $\stackrel{â†’}{\text{AR}}$ â‡’ m ($\stackrel{â†’}{\text{b}}$ - $\stackrel{â†’}{\text{r}}$) = n ( $\stackrel{â†’}{\text{r}}$   - $\stackrel{â†’}{\text{a}}$  ) â‡’ m $\stackrel{â†’}{\text{b}}$ - m $\stackrel{â†’}{\text{r}}$ = n $\stackrel{â†’}{\text{r}}$ - n $\stackrel{â†’}{\text{a}}$ â‡’ m $\stackrel{â†’}{\text{r}}$ + n $\stackrel{â†’}{\text{r}}$= m $\stackrel{â†’}{\text{b}}$ + n $\stackrel{â†’}{\text{a}}$ â‡’ (m + n )$\stackrel{â†’}{\text{r}}$= m $\stackrel{â†’}{\text{b}}$ + n $\stackrel{â†’}{\text{a}}$ â‡’ $\stackrel{â†’}{\text{r}}$=  $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{+ n}\stackrel{â†’}{\text{a}}}{\text{m+n}}$  â‡’ $\stackrel{â†’}{\text{OR}}$ = $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{+ n}\stackrel{â†’}{\text{a}}}{\text{m+n}}$  The point R lies at the midpoint of line segment AB. When m : n = 1:1 â‡’ m = n â‡’ $\stackrel{â†’}{\text{OR}}$ =   $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{+ n}\stackrel{â†’}{\text{a}}}{\text{m+n}}$  =      â‡’ $\stackrel{â†’}{\text{OR}}$ =      $\frac{\text{m (}\stackrel{â†’}{\text{a}}\text{+}\stackrel{â†’}{\text{b}}\text{)}}{\text{2m}}$          â‡’ $\stackrel{â†’}{\text{OR}}$ =    =        Case II: R divides AB externally   $\stackrel{â†’}{\text{OA}}$  =            = $\stackrel{â†’}{\text{b}}$       = $\stackrel{â†’}{\text{r}}$    Let R divide AB externally such that AR:BR = m:n In âˆ†OAR: $\stackrel{â†’}{\text{OR}}$  = $\stackrel{â†’}{\text{OA}}$  + $\stackrel{â†’}{\text{AR}}$  â‡’ $\stackrel{â†’}{\text{AR}}$  = $\stackrel{â†’}{\text{OR}}$  - $\stackrel{â†’}{\text{OA}}$  â‡’ $\stackrel{â†’}{\text{AR}}$  = $\stackrel{â†’}{\text{r}}$  - $\stackrel{â†’}{\text{a}}$  In âˆ†OBR: $\stackrel{â†’}{\text{OR}}$  = $\stackrel{â†’}{\text{OB}}$  + $\stackrel{â†’}{\text{BR}}$  â‡’ $\stackrel{â†’}{\text{BR}}$  = $\stackrel{â†’}{\text{OR}}$  - $\stackrel{â†’}{\text{OB}}$  â‡’ $\stackrel{â†’}{\text{BR}}$  = $\stackrel{â†’}{\text{r}}$  - $\stackrel{â†’}{\text{b}}$  $\stackrel{â†’}{\text{AR}}$  :$\stackrel{â†’}{\text{BR}}$  = m:n â‡’ m ($\stackrel{â†’}{\text{BR}}$ )= n ($\stackrel{â†’}{\text{AR}}$ ) â‡’ m ($\stackrel{â†’}{\text{r}}$  - $\stackrel{â†’}{\text{b}}$ ) = n ($\stackrel{â†’}{\text{r}}$  - $\stackrel{â†’}{\text{a}}$  ) â‡’ m $\stackrel{â†’}{\text{r}}$ - m $\stackrel{â†’}{\text{b}}$ = n $\stackrel{â†’}{\text{r}}$ - n $\stackrel{â†’}{\text{a}}$ â‡’ m $\stackrel{â†’}{\text{r}}$ - n $\stackrel{â†’}{\text{r}}$= m $\stackrel{â†’}{\text{b}}$ - n $\stackrel{â†’}{\text{a}}$ â‡’ (m - n )$\stackrel{â†’}{\text{r}}$ = m $\stackrel{â†’}{\text{b}}$ - n $\stackrel{â†’}{\text{a}}$ â‡’ $\stackrel{â†’}{\text{r}}$ = $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{- n}\stackrel{â†’}{\text{a}}}{\text{m-n}}$  â‡’ $\stackrel{â†’}{\text{OR}}$ = $\frac{\text{m}\stackrel{â†’}{\text{b}}\text{- n}\stackrel{â†’}{\text{a}}}{\text{m-n}}$ .

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