Centre of mass of non symmetric triangle.With a proper derivation and explanation pls

The centroid of a triangle is the point of intersection of its medians (the lines joining each vertex with the midpoint of the opposite side). The centroid divides each of the medians in the ratio 2:1, which is to say it is located ⅓ of the distance from each side to the opposite vertex (see figures at right). Its Cartesian coordinates are the means of the coordinates of the three vertices. That is, if the three vertices are {\displaystyle L=(x_{L},y_{L}),} {\displaystyle M=(x_{M},y_{M}),} and {\displaystyle N=(x_{N},y_{N}),} then the centroid (denoted C here but most commonly denoted G in triangle geometry) is

{\displaystyle C={\frac {1}{3}}(L+M+N)=\left({\frac {1}{3}}(x_{L}+x_{M}+x_{N}),\;\;{\frac {1}{3}}(y_{L}+y_{M}+y_{N})\right).}

The centroid is therefore at {\displaystyle {\tfrac {1}{3}}:{\tfrac {1}{3}}:{\tfrac {1}{3}}} in barycentric coordinates.

In trilinear coordinates the centroid can be expressed in any of these equivalent ways in terms of the side lengths a, b, cand vertex angles L, M, N:

{\displaystyle =\sec L+\sec M\cdot \sec N:\sec M+\sec N\cdot \sec L:\sec N+\sec L\cdot \sec M.}

{\displaystyle =\cos L+\cos M\cdot \cos N:\cos M+\cos N\cdot \cos L:\cos N+\cos L\cdot \cos M}

{\displaystyle C={\frac {1}{a}}:{\frac {1}{b}}:{\frac {1}{c}}=bc:ca:ab=\csc L:\csc M:\csc N}

The centroid is also the physical center of mass if the triangle is made from a uniform sheet of material; or if all the mass is concentrated at the three vertices, and evenly divided among them. On the other hand, if the mass is distributed along the triangle's perimeter, with uniform linear density, then the center of mass lies at the Spieker center (the incenter of the medial triangle), which does not (in general) coincide with the geometric centroid of the full triangle.

The area of the triangle is 1.5 times the length of any side times the perpendicular distance from the side to the centroid.

A triangle's centroid lies on its Euler line between its orthocenter H and its circumcenter O, exactly twice as close to the latter as to the former:

{\displaystyle CH=2CO.}

In addition, for the incenter I and nine-point center N, we have

{\displaystyle CH=4CN,}

{\displaystyle CO=2CN,}

{\displaystyle IC

{\displaystyle IH

{\displaystyle IC

If G is the centroid of the triangle ABC, then:

• {\displaystyle \displaystyle ({\text{Area of triangle ABG}})=({\text{Area of triangle ACG}})=({\text{Area of triangle BCG}})={\frac {1}{3}}({\text{Area of triangle ABC}})}

The isogonal conjugate of a triangle's centroid is its symmedian point.

i hope it helped u.

Approved