Section formula

In coordinate geometry, the Section formula is a formula used to find the ratio in which a line segment is divided by a point internally or externally.^[1] It is used to find out the centroid, incenter and excenters of a triangle. In physics, it is used to find the center of mass of systems, equilibrium points, etc.^[2][3][4][5]

Internal Divisions

Internal division with section formula

If point P (lying on AB) divides the line segment AB joining the points ${\displaystyle \mathrm {A} (x_{1},y_{1})}$ and ${\displaystyle \mathrm {B} (x_{2},y_{2})}$ in the ratio m:n, then

${\displaystyle P=\left({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}}\right)}$^[6]

The ratio m:n can also be written as ${\displaystyle m/n:1}$, or ${\displaystyle k:1}$, where ${\displaystyle k=m/n}$. So, the coordinates of point ${\displaystyle P}$ dividing the line segment joining the points ${\displaystyle \mathrm {A} (x_{1},y_{1})}$ and ${\displaystyle \mathrm {B} (x_{2},y_{2})}$ are:

${\displaystyle \left({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}}\right)}$

${\displaystyle =\left({\frac {{\frac {m}{n}}x_{2}+x_{1}}{{\frac {m}{n}}+1}},{\frac {{\frac {m}{n}}y_{2}+y_{1}}{{\frac {m}{n}}+1}}\right)}$

${\displaystyle =\left({\frac {kx_{2}+x_{1}}{k+1}},{\frac {ky_{2}+y_{1}}{k+1}}\right)}$^[4][5]

Similarly, the ratio can also be written as ${\displaystyle k:(1-k)}$, and the coordinates of P are ${\displaystyle ((1-k)x_{1}+kx_{2},(1-k)y_{1}+ky_{2})}$.^[1]

Proof

Triangles ${\displaystyle PAQ\sim BPC}$.

${\displaystyle {\begin{aligned}{\frac {AP}{BP}}={\frac {AQ}{CP}}={\frac {PQ}{BC}}\\{\frac {m}{n}}={\frac {x-x_{1}}{x_{2}-x}}={\frac {y-y_{1}}{y_{2}-y}}\\mx_{2}-mx=nx-nx_{1},my_{2}-my=ny-ny_{1}\\mx+nx=mx_{2}+nx_{1},my+ny=my_{2}+ny_{1}\\(m+n)x=mx_{2}+nx_{1},(m+n)y=my_{2}+ny_{1}\\x={\frac {mx_{2}+nx_{1}}{m+n}},y={\frac {my_{2}+ny_{1}}{m+n}}\\\end{aligned}}}$

External Divisions

External division with section formula

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

${\displaystyle P=\left({\dfrac {mx_{2}-nx_{1}}{m-n}},{\dfrac {my_{2}-ny_{1}}{m-n}}\right)}$^[6]

Proof

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

${\displaystyle {\begin{aligned}{\frac {AB}{BP}}={\frac {AC}{BD}}={\frac {PC}{PD}}\\{\frac {m}{n}}={\frac {x-x_{1}}{x-x_{2}}}={\frac {y-y_{1}}{y-y_{2}}}\\mx-mx_{2}=nx-nx_{1},my-my_{2}=ny-ny_{1}\\mx-nx=mx_{2}-nx_{1},my-ny=my_{2}-ny_{1}\\(m-n)x=mx_{2}-nx_{1},(m-n)y=my_{2}-ny_{1}\\x={\frac {mx_{2}-nx_{1}}{m-n}},y={\frac {my_{2}-ny_{1}}{m-n}}\\\end{aligned}}}$

Midpoint formula

Main article: Midpoint

The midpoint of a line segment divides it internally in the ratio ${\textstyle 1:1}$. Applying the Section formula for internal division:^[4][5]

$${\displaystyle P=\left({\dfrac {x_{1}+x_{2}}{2}},{\dfrac {y_{1}+y_{2}}{2}}\right)}$$

Derivation

${\displaystyle P=\left({\dfrac {mx_{2}+nx_{1}}{m+n}},{\dfrac {my_{2}+ny_{1}}{m+n}}\right)}$

${\displaystyle =\left({\frac {1\cdot x_{1}+1\cdot x_{2}}{1+1}},{\frac {1\cdot y_{1}+1\cdot y_{2}}{1+1}}\right)}$

${\displaystyle =\left({\dfrac {x_{1}+x_{2}}{2}},{\dfrac {y_{1}+y_{2}}{2}}\right)}$

Centroid

Centroid of a triangle

The centroid of a triangle is the intersection of the medians and divides each median in the ratio ${\textstyle 2:1}$. Let the vertices of the triangle be ${\displaystyle A(x_{1},y_{1})}$, ${\textstyle B(x_{2},y_{2})}$ and ${\textstyle C(x_{3},y_{3})}$. So, a median from point A will intersect BC at ${\textstyle \left({\frac {x_{2}+x_{3}}{2}},{\frac {y_{2}+y_{3}}{2}}\right)}$. 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 3-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

${\displaystyle \left({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}},{\frac {mz_{2}+nz_{1}}{m+n}}\right)}$^[7]

If, instead, P is a point on the line such that ${\displaystyle AP:PB=k:1-k}$, its coordinates are ${\displaystyle ((1-k)x_{1}+kx_{2},(1-k)y_{1}+ky_{2},(1-k)z_{1}+kz_{2})}$.^[7]

In vectors

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

1. in the ratio ${\displaystyle m:n}$ internally, is given by ${\displaystyle {\frac {n{\vec {a}}+m{\vec {b}}}{m+n}}}$^[8][1]
2. in the ratio ${\displaystyle m:n}$ externally, is given by ${\displaystyle {\frac {m{\vec {b}}-n{\vec {a}}}{m-n}}}$^[8]

See also

• Cross-section Formula
• Distance Formula
• Midpoint Formula