Section Formula

Category : 9th Class

Section Formula

Let the coordinate of end points of a line segment be $({{x}_{1}},{{y}_{1}})$and$({{x}_{2}},{{y}_{2}})$ then the coordinate of point C which divides the line segment in the ratio m : n is:

$x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\,\,y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$

Proof:

Let point $C(x,y)$ be the point which divides the line segment joining two given points  $A({{x}_{1}},{{y}_{1}})$ and $B({{x}_{2}},{{y}_{2}})$ internally in the ratio m : n.

From A, B and C draw AD, BF and CE perpendicular to x - axis.

From A and C draw AG and CH perpendicular to CE and BF.

Then $AG={{x}_{1}}-{{x}_{1}}$

$CH={{x}_{2}}-x$

$CG=y-{{y}_{1}}$

$BH={{y}_{2}}-y$

$\Delta \text{AGC}$ and $\Delta \text{BCH}$ are similar. Therefore,

$\frac{AG}{CH}=\frac{CG}{BH}=\frac{AC}{BC}$

$\Rightarrow$ $\frac{x-{{x}_{1}}}{{{x}_{2}}-x}=\frac{y-{{y}_{1}}}{{{y}_{2}}-y}=\frac{m}{n}$ $\Rightarrow$ $\frac{x-{{x}_{1}}}{{{x}_{2}}-x}=\frac{m}{n}$ $\Rightarrow$ $nx-n{{x}_{1}}=m{{x}_{2}}-mx$ $\Rightarrow$

$(m+n)x=m{{x}_{2}}+n{{x}_{1}}$                 $x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n}$

Similarly $\frac{y-{{y}_{1}}}{{{y}_{2}}-y}=\frac{m}{n}$    $\Rightarrow$                $y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$

Thus the coordinate of point C is$\left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$

Note: If point C is the midpoint of AB then the coordinate of C is $\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)$

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