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.

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. P = \left(\frac{mx_2 + nx_1}{m + n},\frac{my_2 + ny_1}{m + n}\right) 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 \mathrm{A}(x_1,y_1) and \mathrm{B}(x_2,y_2) are: \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right) =\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 ) =\left ( \frac{kx_2 +x_1}{k +1},\frac{ky_2 + y_1}{k +1} \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) . \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} P = \left(\dfrac{mx_2 - nx_1}{m - n}, \dfrac{my_2 - ny_1}{m - n}\right) Therefore ∠ACP = ∠BDP \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} The midpoint of a line segment divides it internally in the ratio 1:1 . Applying the Section formula for internal division: P = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) P = \left(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\right) = \left ( \frac{1\cdot x_1 + 1\cdot x_2}{1+1},\frac{1 \cdot y_1 + 1\cdot y_2}{1+1} \right ) =\left(\dfrac{x_1 + x_2}{2}, \dfrac{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 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: \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 (x 1 , y 1 , z 1 ) and (x 2 , y 2 , z 2 ) 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 \left ( \frac{mx_2 + nx_1}{m +n} ,\frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right ) 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) .

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