School of Mathematics and Statistics, The University of Sydney
 5. Division of a line segment
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Internal division of a line segment

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With an assigned point O as origin, the position of any point P is given uniquely by the vector ---> OP, which is called the position vector of P relative to O.

Let P1 and P2 be any points, and let R be a point on the line P1P2 such that R divides the line segment P1P2 in the ratio m : n. That is, R is the point such that ----> P1R = mn-----> RP2. Our task is to find the position vector of R (relative to O) in terms of the position vectors of P1 and P2.

  P1                       R                                P2     O

As ----> P1R = m-  n----> RP2, we have n----> P1R = m----> RP2 and therefore

   --->    ---->        ---->    ---> n( OR  - OP1)  = m( OP2 -  OR),
(1)

which rearranges to give

        ---->      ----> --->     nOP1  + m OP2 OR  =  ---------------,    m +  n /= 0.            m  + n

When m and n are both positive, the vectors ----> P1R and ----> RP2 have the same direction, since ----> P1R = m- n----> RP2. This corresponds to the situation where R lies between P1 and P2, as shown in the diagram above. R is then said to divide the line segment P1P2 internally in the ratio m : n.

As a special case of the general formula (1) we can obtain a formula for the position vector ----> OM of the midpoint M between two points P1 and P2.

In this case, m : n = 1 : 1 and so

---->     1( ---->    ---->  ) OM   =  2  OP1 + OP2   .

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