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Let P be the point that divides the line segment AB in the ration AP:PB=m:n.
If O is the origin show that
OP=(n/m+n)OA + (m/m+n)OB
Dear student,
Solution:
Step 1: Assume that the point P is dividing the line segment AB in the ratio m: n initially.
Step 2: Assume that P is not unique.
Step 3: Then there is another point Q on AB which divides AB in the ratio m: n internally.
From the above diagram we understand
= and =
n AP = m PB and n AQ = m QB
n AP = m(AB – AP) and n AQ = m (AB – AQ)
n AP = m AB – m AP and n AQ = m AB – m AQ
n AP + m AP = m AB and n AQ + m AQ = m AB
(m + n) AP = m AB and (m + n) AQ = m AB
AP = , AQ =
From above derivation we get AP = AQ
Therefore P and Q are the same point. It is opposite to our assumption. Therefore given (if P is a point which partitions the line segment AB in the ratio m: n, than P is unique) is proved.
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