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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 thatOP=(n/m+n)OA + (m/m+n)OB

ilham rafie , 13 Years ago

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1 Answers

SAGAR SINGH - IIT DELHI

Last Activity: 13 Years ago

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

$\displaystyle\frac{{{A}{P}}}{{{P}{B}}}$ = $\displaystyle\frac{{{m}}}{{{n}}}$ and $\displaystyle\frac{{{A}{Q}}}{{{Q}{P}}}$ = $\displaystyle\frac{{{m}}}{{{n}}}$

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 = $\displaystyle\frac{{{m}{A}{P}}}{{{m}+{n}}}$ , AQ = $\displaystyle\frac{{{m}{A}{P}}}{{{m}+{n}}}$

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.