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# Section Formula

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#### Section Formula - Lesson Summary

The position vector for R in both these cases.

Case I: R divides AB internally.

OA â†’   =  a â†’

OB â†’   =  b â†’

OR â†’   =  r â†’

Point R divide line segment AB in the ratio m:n

AR:RB = m:n

In âˆ†OAR:

OR â†’    = OA â†’   +  AR â†’

â‡’ AR â†’ = OR â†’ - OA â†’

â‡’ AR â†’ = r â†’ - a â†’

In âˆ†ORB:

OB â†’ = OR â†’ + RB â†’

â‡’ RB â†’ = OB â†’ - OR â†’

â‡’ RB â†’ = b â†’ - r â†’

AR â†’ :RB â†’ = m:n

â‡’ m RB â†’ = n AR â†’

â‡’ m (b â†’ - r â†’ ) = n ( r â†’    - a â†’   )

â‡’ m b â†’ - m r â†’ = n r â†’ - n a â†’

â‡’ m r â†’ + n r â†’ = m b â†’ + n a â†’

â‡’ (m + n )r â†’ = m b â†’ + n a â†’

â‡’ r â†’ =  m b â†’ + n a â†’ m+n

â‡’ OR â†’ = m b â†’ + n a â†’ m+n

The point R lies at the midpoint of line segment AB.

When m : n = 1:1

â‡’ m = n

â‡’ OR â†’ =   m b â†’ + n a â†’ m+n   =    m b â†’ + m a â†’  m+m

â‡’ OR â†’ =       m ( a â†’ + b â†’ ) 2m

â‡’ OR â†’ =   ( b â†’ + a â†’ ) 2   =  ( a â†’ + b â†’ ) 2

Case II: R divides AB externally

OA â†’   =  a â†’

OB â†’   =  b â†’

OR â†’   =  r â†’

Let R divide AB externally such that AR:BR = m:n

In âˆ†OAR:

OR â†’   = OA â†’   + AR â†’

â‡’ AR â†’   = OR â†’   - OA â†’

â‡’ AR â†’   = r â†’   - a â†’

In âˆ†OBR:

OR â†’   = OB â†’   + BR â†’

â‡’ BR â†’   = OR â†’   - OB â†’

â‡’ BR â†’   = r â†’   - b â†’

AR â†’   :BR â†’   = m:n

â‡’ m (BR â†’  )= n (AR â†’  )

â‡’ m (r â†’   - b â†’  ) = n (r â†’   - a â†’   )

â‡’ m r â†’ - m b â†’ = n r â†’ - n a â†’

â‡’ m r â†’ - n r â†’ = m b â†’ - n a â†’

â‡’ (m - n )r â†’ = m b â†’ - n a â†’

â‡’ r â†’ = m b â†’ - n a â†’ m-n

â‡’ OR â†’ = m b â†’ - n a â†’ m-n  .