Section formula

File:Internal division.png

If point P (lying on AB) divides the line segment AB joining the points $\backslash mathrm\{A\}(x\_1,y\_1)$ and $\backslash mathrm\{B\}(x\_2,y\_2)$ in the ratio m:n, then

$P\; =\; \backslash left(\backslash frac\{mx\_2\; +\; nx\_1\}\{m\; +\; n\},\backslash frac\{my\_2\; +\; ny\_1\}\{m\; +\; n\}\backslash right)${{Cite book|last=Loney|first=S L|title=The Elements of Coordinate Geometry (Part-1)}}

The ratio m:n can also be written as $m/n:1$, or $k:1$, where $k=m/n$. So, the coordinates of point $P$ dividing the line segment joining the points $\backslash mathrm\{A\}(x\_1,y\_1)$ and $\backslash mathrm\{B\}(x\_2,y\_2)$ are:

$\backslash left(\backslash frac\{mx\_2\; +\; nx\_1\}\{m\; +\; n\},\; \backslash frac\{my\_2\; +\; ny\_1\}\{m\; +\; n\}\backslash right)$

$=\backslash left(\backslash frac\{\backslash frac\{m\}\{n\}x\_2\; +x\_1\}\{\backslash frac\{m\}\{n\}+1\},\backslash frac\{\backslash frac\{m\}\{n\}y\_2\; +y\_1\}\{\backslash frac\{m\}\{n\}+1\}\; \backslash right\; )$

$=\backslash left\; (\; \backslash frac\{kx\_2\; +x\_1\}\{k\; +1\},\backslash frac\{ky\_2\; +\; y\_1\}\{k\; +1\}\; \backslash right\; )$

Similarly, the ratio can also be written as $k:(1-k)$, and the coordinates of P are $((1-k)x\_1\; +\; kx\_2,\; (1-k)y\_1\; +\; ky\_2)$.

Triangles $PAQ\backslash sim\; BPC$.

:$\backslash begin\{align\}$

\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}

File:External division.png

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

$P\; =\; \backslash left(\backslash dfrac\{mx\_2\; -\; nx\_1\}\{m\; -\; n\},\; \backslash dfrac\{my\_2\; -\; ny\_1\}\{m\; -\; n\}\backslash right)$

Triangles $PAC\backslash sim\; PBD$ (Let C and D be two points where A & P and B & P intersect respectively).

Therefore ∠ACP = ∠BDP

:$\backslash begin\{align\}$

\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}

{{Main articles|Midpoint}}

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

$$P\; =\; \backslash left(\backslash dfrac\{x\_1\; +\; x\_2\}\{2\},\; \backslash dfrac\{y\_1\; +\; y\_2\}\{2\}\; \backslash right)$$

$P\; =\; \backslash left(\backslash dfrac\{mx\_2\; +\; nx\_1\}\{m\; +\; n\},\; \backslash dfrac\{my\_2\; +\; ny\_1\}\{m\; +\; n\}\backslash right)$

$=\; \backslash left\; (\; \backslash frac\{1\backslash cdot\; x\_1\; +\; 1\backslash cdot\; x\_2\}\{1+1\},\backslash frac\{1\; \backslash cdot\; y\_1\; +\; 1\backslash cdot\; y\_2\}\{1+1\}\; \backslash right\; )$

$=\backslash left(\backslash dfrac\{x\_1\; +\; x\_2\}\{2\},\; \backslash dfrac\{y\_1\; +\; y\_2\}\{2\}\; \backslash right)$

File:Centroid.svg

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 $A(x\_1,\; y\_1)$, $B(x\_2,\; y\_2)$ and $C(x\_3,\; y\_3)$. So, a median from point A will intersect BC at $\backslash left(\backslash frac\{x\_2\; +\; x\_3\}\{2\},\; \backslash frac\{y\_2\; +\; y\_3\}\{2\}\backslash right)$.

Using the section formula, the centroid becomes:

:$$$$

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

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 $AP:PB\; =m:n$. Then the section formula gives the coordinates of P as

$\backslash left\; (\; \backslash frac\{mx\_2\; +\; nx\_1\}\{m\; +n\}\; ,\backslash frac\{my\_2\; +\; ny\_1\}\{m+n\},\; \backslash frac\{mz\_2\; +\; nz\_1\}\{m+n\}\; \backslash right\; )${{Citation|last1=Clapham|first1=Christopher|title=section formulae|date=2014-09-18|url=https://www.oxfordreference.com/view/10.1093/acref/9780199679591.001.0001/acref-9780199679591-e-2547|work=The Concise Oxford Dictionary of Mathematics|publisher=Oxford University Press|language=en|doi=10.1093/acref/9780199679591.001.0001|isbn=978-0-19-967959-1|access-date=2020-10-30|last2=Nicholson|first2=James|url-access=subscription}}

If, instead, P is a point on the line such that $AP:PB\; =\; k:1-k$, its coordinates are $((1-k)x\_1\; +\; kx\_2,\; (1-k)y\_1\; +\; ky\_2,\; (1-k)z\_1\; +\; kz\_2)$.

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

1. in the ratio $m:n$ internally, is given by $\backslash frac\{n\backslash vec\{a\}\; +\; m\backslash vec\{b\}\}\{m+n\}${{Cite web| title=Vector Algebra | url=https://ncert.nic.in/ncerts/l/leep210.pdf | archive-url=https://web.archive.org/web/20161213040342/http://ncert.nic.in:80/ncerts/l/leep210.pdf | archive-date=2016-12-13}}
2. in the ratio $m:n$ externally, is given by $\backslash frac\{m\backslash vec\{b\}\; -\; n\backslash vec\{a\}\}\{m-n\}$

• Cross-section Formula
• Distance Formula
• Midpoint Formula

{{Reflist}}

• [https://help.geogebra.org/topic/section-formula section-formula] by GeoGebra

Category:Analytic geometry