विभाजन-सूत्र

The Section formula is used to find the coordinates of the point that divides a line segment (externally or internally) into some ratio.

Derivation of Section Formula

Fig 1 - Section Formula

Consider any two points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ and assume that $P(x,y)$ divides $AB$ internally in the ratio $m_{1}:m_{2}$ ,

i.e., ${\frac {PA}{PB}}={\frac {m_{1}}{m_{2}}}$ (see Fig. 1).

Draw $AC,PD,BE$ perpendicular to the $x-$axis. Draw $AR,PQ$ parallel to the $x-$axis. Then, by the $AA$ similarity criterion,

$\bigtriangleup PAR\sim \bigtriangleup BPQ$

${\frac {PA}{PB}}={\frac {AR}{PQ}}={\frac {PR}{BQ}}.......(1)$

$AR=CD=OD-OC=x-x_{1}$

$PQ=DE=OE-OD=x_{2}-x$

$PR=PD-RD=PD-AC=y-y_{1}$

$BQ=BE-QE=BE-PD=y_{2}-y$

Substituting in $(1)$ we get

${\frac {m_{1}}{m_{2}}}={\frac {x-x_{1}}{x_{2}-x}}={\frac {y-y_{1}}{y_{2}-y}}$

Taking ${\frac {m_{1}}{m_{2}}}={\frac {x-x_{1}}{x_{2}-x}}$

$m_{1}x_{2}-m_{1}x=m_{2}x-m_{2}x_{1}$

$m_{1}x_{2}+m_{2}x_{1}=m_{2}x+m_{1}x$

$m_{1}x_{2}+m_{2}x_{1}=x(m_{2}+m_{1})$

$x={\frac {m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}}$

Taking ${\frac {m_{1}}{m_{2}}}={\frac {y-y_{1}}{y_{2}-y}}$

$m_{1}y_{2}-m_{1}y=m_{2}y-m_{2}y_{1}$

$m_{1}y_{2}+m_{2}y_{1}=m_{2}y+m_{1}y$

$m_{1}y_{2}+m_{2}y_{1}=y(m_{2}+m_{1})$

$y={\frac {m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}}$

So, the coordinates of the point $P(x,y)$ which divides the line segment joining the points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$, internally, in the ratio $m_{1}:m_{2}$ are $\left({\frac {m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}},{\frac {m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}}\right)$ is the section formula.

Example

Find the coordinates of the point which divides the line segment joining the points $(4,-3)$ and $(8,5)$ in the ratio $3:1$ internally.

Solution : Let $P(x,y)$ be the required point.

$x_{1}=4,y_{1}=-3$

$x_{2}=8,y_{2}=5$

$m_{1}=3,m_{2}=1$

Using the section formula, we get

$x={\frac {m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}}$

$x={\frac {3\times 8+1\times 4}{3+1}}=7$

$y={\frac {m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}}$

$y={\frac {3\times 5+1\times -3}{3+1}}=3$

Therefore $(7,3)$ is the required point.