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. 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.

Internal Divisions

Internal division with section formula

If point P (lying on AB) divides the line segment AB joining the points [math]\displaystyle{ \mathrm{A}(x_1,y_1) }[/math] and [math]\displaystyle{ \mathrm{B}(x_2,y_2) }[/math] in the ratio m:n, then

[math]\displaystyle{ P = \left(\frac{mx_2 + nx_1}{m + n},\frac{my_2 + ny_1}{m + n}\right) }[/math]

The ratio m:n can also be written as [math]\displaystyle{ m/n:1 }[/math], or [math]\displaystyle{ k:1 }[/math], where [math]\displaystyle{ k=m/n }[/math]. So, the coordinates of point [math]\displaystyle{ P }[/math] dividing the line segment joining the points [math]\displaystyle{ \mathrm{A}(x_1,y_1) }[/math] and [math]\displaystyle{ \mathrm{B}(x_2,y_2) }[/math] are:

[math]\displaystyle{ \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right) }[/math]

[math]\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 ) }[/math]

[math]\displaystyle{ =\left ( \frac{kx_2 +x_1}{k +1},\frac{ky_2 + y_1}{k +1} \right ) }[/math]

Similarly, the ratio can also be written as [math]\displaystyle{ k:k-1 }[/math], and the coordinates of P are [math]\displaystyle{ ((1-k)x_1 + kx_2, (1-k)y_1 + ky_2) }[/math].

Proof

Triangles [math]\displaystyle{ PAQ\sim BPC }[/math].

[math]\displaystyle{ \begin{align} \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{align} }[/math]

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

[math]\displaystyle{ P = \left(\dfrac{mx_2 - nx_1}{m - n}, \dfrac{my_2 - ny_1}{m - n}\right) }[/math]

Proof

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

[math]\displaystyle{ \begin{align} \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{align} }[/math]

Midpoint formula

The midpoint of a line segment divides it internally in the ratio [math]\displaystyle{ 1:1 }[/math]. Applying the Section formula for internal division:

[math]\displaystyle{ P = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) }[/math]

Derivation

[math]\displaystyle{ P = \left(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\right) }[/math]

[math]\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 ) }[/math]

[math]\displaystyle{ =\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) }[/math]

Centroid

Centroid of a triangle

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

[math]\displaystyle{ \left(\frac{x_1 + x_2 + x_3}{3},\frac{y_1 + y_2 + y_3}{3} \right) }[/math]

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 [math]\displaystyle{ AP:PB =m:n }[/math]. Then the section formulae give the coordinates of P as

[math]\displaystyle{ \left ( \frac{mx_2 + nx_1}{m +n} ,\frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right ) }[/math]

If, instead, P is a point on the line such that [math]\displaystyle{ AP:PB = k:1-k }[/math], its coordinates are [math]\displaystyle{ ((1-k)x_1 + kx_2, (1-k)y_1 + ky_2, (1-k)z_1 + kz_2) }[/math].

In vectors

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

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

See also

• Cross-section Formula
• Distance Formula
• Midpoint Formula

External links

• section-formula by GeoGebra