Section formula

External divisions

If a point P (lying on the extension of AB) divides AB in the ratio m:n then

${\displaystyle P=\left({\displaystyle \frac{m{x}_2-n{x}_1}{m-n}},{\displaystyle \frac{m{y}_2-n{y}_1}{m-n}}\right)}$

Proof

Triangles ${\displaystyle PAC\sim PBD}$ (Let C and D be two points where A & P and B & P intersect respectively). Therefore ∠ACP = ∠BDP

${\displaystyle \begin{array}{c}\frac{AB}{BP}=\frac{AC}{BD}=\frac{PC}{PD}\\ \frac{m}{n}=\frac{x-x_1}{x-x_2}=\frac{y-y_1}{y-y_2}\\ mx-m{x}_2=nx-n{x}_1,my-m{y}_2=ny-n{y}_1\\ mx-nx=m{x}_2-n{x}_1,my-ny=m{y}_2-n{y}_1\\ (m-n)x=m{x}_2-n{x}_1,(m-n)y=m{y}_2-n{y}_1\\ x=\frac{m{x}_2-n{x}_1}{m-n},y=\frac{m{y}_2-n{y}_1}{m-n}\end{array}}$

Midpoint formula

The midpoint of a line segment divides it internally in the ratio $1:1$. Applying the Section formula for internal division:

$${\displaystyle P=\left({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}}\right)}$$

Derivation

${\displaystyle P=\left({\displaystyle \frac{m{x}_2+n{x}_1}{m+n}},{\displaystyle \frac{m{y}_2+n{y}_1}{m+n}}\right)}$

${\displaystyle =\left(\frac{1\cdot x_1+1\cdot x_2}{1+1},\frac{1\cdot y_1+1\cdot y_2}{1+1}\right)}$

${\displaystyle =\left({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}}\right)}$

Centroid

The centroid of a triangle is the intersection of the medians and divides each median in the ratio $2:1$. Let the vertices of the triangle be ${\displaystyle A(x_1,y_1)}$, $B(x_2,y_2)$ and $C(x_3,y_3)$. So, a median from point A will intersect BC at $\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$. Using the section formula, the centroid becomes:

$${\displaystyle \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)}$$

In three dimensions

Let A and B be two points with Cartesian coordinates (x₁, y₁, z₁) and (x₂, y₂, z₂) and P be a point on the line through A and B. If ${\displaystyle AP:PB=m:n}$. Then the section formula gives the coordinates of P as

${\displaystyle \left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n}\right)}$

If, instead, P is a point on the line such that ${\displaystyle AP:PB=k:1-k}$, its coordinates are ${\displaystyle ((1-k)x_1+k{x}_2,(1-k)y_1+k{y}_2,(1-k)z_1+k{z}_2)}$.

In vectors

The position vector of a point P dividing the line segment joining the points A and B whose position vectors are ${\displaystyle \overrightarrow{a}}$ and ${\displaystyle \overrightarrow{b}}$

1. in the ratio ${\displaystyle m:n}$ internally, is given by ${\displaystyle \frac{n\overrightarrow{a}+m\overrightarrow{b}}{m+n}}$
2. in the ratio ${\displaystyle m:n}$ externally, is given by ${\displaystyle \frac{m\overrightarrow{b}-n\overrightarrow{a}}{m-n}}$