Section Formula (Internal Division)

Intro: Compute the coordinates of the point dividing the segment in ratio m:n.

Worked example

• Find point dividing $A(-1,4)$, $B(9,-6)$ in ratio $m:n=2:3$ (from $A$ to $B$).



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• Given: $A(x_1,y_1)=(-1,4)$, $B(x_2,y_2)=(9,-6)$, internal ratio $m:n=2:3$ (measured from $A$ toward $B$).



• Section formula (internal): $$x=\frac{m x_2 + n x_1}{m+n},\quad y=\frac{m y_2 + n y_1}{m+n},\qquad m,n>0.$$



• Compute $x$-coordinate: $$x=\frac{2\cdot 9 + 3\cdot(-1)}{2+3}=\frac{18-3}{5}=\frac{15}{5}=3.$$



• Compute $y$-coordinate: $$y=\frac{2\cdot(-6) + 3\cdot 4}{2+3}=\frac{-12+12}{5}=\frac{0}{5}=0.$$



• Therefore: $$\boxed{P(3,0)}.$$



• Vector (parametric) check: $$\overrightarrow{AB}=B-A=(9-(-1),\,-6-4)=(10,-10).$$ With $\tfrac{m}{m+n}=\tfrac{2}{5}$, $$A+\tfrac{2}{5}\,\overrightarrow{AB}=(-1,4)+\tfrac{2}{5}(10,-10)=(-1,4)+(4,-4)=(3,0).$$



• Ratio check along each axis: $$\frac{AP_x}{PB_x}=\frac{|3-(-1)|}{|9-3|}=\frac{4}{6}=\frac{2}{3},\qquad \frac{AP_y}{PB_y}=\frac{|0-4|}{|-6-0|}=\frac{4}{6}=\frac{2}{3}.$$ Both match $\tfrac{m}{n}=\tfrac{2}{3}$.



• Note (external division): using $$x=\frac{m x_2 - n x_1}{m-n},\quad y=\frac{m y_2 - n y_1}{m-n}$$ when the point divides $AB$ externally in the ratio $m:n$ (one of $m,n$ effectively negative).

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Related

• Midpoint $(m=n=1)$