Centroid - Definition, Properties, Theorem and Formulas

A centroid is a term used in coordinate geometry to represent the center of the mathematical object. In this article, we will learn more about the definition of centroid and centroid formulas for all shapes.

The topic of centroid falls under category of coordinate geometry, is a crucial chapter in the syllabus of class 11th mathematics. It is equally important in terms of both board and competitive level exams such as JEE Main, SRMJEE, BITSAT, etc. A total of 17 questions were asked in the paper of JEE Mains (2013-2023) from the same topic.

Centroid of Different Shapes

Centroid of different shapes includes the centroid of triangle, semicircle, square and trapezium.

Before looking into the centroid formula of different shapes, let us see what is centroid.

A centroid is a term used to represent the center of any mathematical object.

What is Centroid of a Triangle?

A triangle is a three sided shape. It has three types, namely, equilateral trianlge, isosceles triangle and scalene trianlge.

Centroid definition for centroid of triangle: A centroid can be defined as the intersection point of the medians of a triangle whereas a median refers to a line joining the mid points of a side and opposite vertex of a triangle. It divides the median of a triangle in the ratio of $2:1$.

Let us look into the centroid of triangle formula in detail.

Centroid of equilateral triangle

We can calculate the coordinates of centroid of triangle by using the formula discussed below. It can be calculated by application of section formula of a line.

Let $\mathrm{A}(x_1,y_1)$, $B(x_2,y_2)$, and $C(x_3,y_3)$ be the vertices of a triangle whose AD, BE and CF. So, D, E, and F are, respectively, the midpoints of BC, CA, and AB.

Centroid of Triangle Formula: The coordinates of the centroid of a triangle (G) whose vertices are $\mathrm{A}({\mathrm{x}}_1$,$y_1)$, $\mathrm{B}({\mathrm{x}}_2,{\mathrm{y}}_2)$, and $C(x_3,y_3)$ are given by:

$(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3}{3})$

Note:

If $D(a_1,b_1)$, $E(a_2,b_2)$, and $F(a_3,b_3)$ are the midpoint of $\mathrm{△}\mathrm{ABC}$, then the centroid of triangle ABC is given by

$(\frac{a_1+a_2+a_3}{3},\frac{b_1+b_2+b_3}{3})$

For example: If there is a right angled triangle having vertices $A(0,0)$, $B(0,b)$, and $C(a,0)$, the coordinates of centroid is given by:

$G(\frac{3\cdot 0+0+a}{3},\frac{3\cdot 0+b+0}{3})=(\frac{3a}{3},\frac{3b}{3})$

Centroid of semicircle

The centroid of semicircle $(x,y)$ is $(r,\frac{4r}{3\pi })$ where $r$ is the radius of the semicircle.

Centroid of a Square

The point where the diagonals of the square intersect each other is the centroid of the square. Same as the triangle, centroid of square is the center of the shape, it is located at intersection of diagonals of the square.

Properties of Centroid of Square

• It divides the square into four equal areas.
• It can easily be calculated by finding the coordinate of midpoint of the diagonal, due to the symmetrical properties of the square.

Centroid of Trapezium

We know that a trapezium is a quadrilateral with two parallel sides. Hence, the centroid of a trapezium lies between its two bases. The formula to calculate the coordinates of centroid of a trapezium is given by, (a,b replace by p,q)

$G=(\frac{h}{2},\frac{(q+2p)}{3(p+b)h})$

here,

$p,q$ denote the length of the parallel sides

$h$ denotes the distance between the parallel sides

Difference between centroid and centre of gravity is that centroid is the geometric center of an object while centre of gravity is that point where the entire mass of an object gets concentrated.

Properties of centroid

The properties of centroid include,

• Centroid is the ‘center of object’ or can even be called as ‘center of gravity’.
• Centroid divides each of the medians of triangle into the ratio of $2:1$.
• It always lies inside the object.
• The Centroid of the triangle and that of a triangle formed by midpoints of sides are the same.
• The three medians of a triangle divide the triangle into six equal areas.
• In a triangle, the line segments connecting the midpoints of sides are parallel to the opposite sides. The four triangles created by these line segments are congruent and thus have equal areas.

Now, let us look into some example of centroid.

Solved Examples Based on Centroid

Example 1: The equations of the sides AB, BC, and CA of a triangle ABC are: $2x+y=0,x+py=21a,(a\ne 0)$ and $x-y=3$ respectively. Let $P(2,a)$ be the centroid of $\mathrm{△}ABC$ Then $(BC{)}^2$ is equal to [JEE MAINS 2023]

Solution

$\begin{array}{rl}\frac{\alpha +\beta +4}{3}=2& \frac{-2\alpha -2+\beta }{3}=a\\ \alpha +\beta =2& -2\alpha +\beta =3\mathrm{a}+2\\ & -2\alpha +2-\alpha =3\mathrm{a}+2\\ & \alpha =-\mathrm{a}\end{array}$

put 'B' in BC

$\begin{array}{rl}& \alpha -2p\alpha =21a\\ & \alpha \cdot (1-2p)=21\mathrm{a}2p-1=21\end{array}$

$p=11$

put 'C' in BC

$\begin{array}{rl}& \beta +3+11\beta =21a\\ & 21\alpha +12\beta +3=0\\ & \beta =2-\alpha \end{array}$
Solving $\alpha =-3,\beta =5$

$\begin{array}{rl}\therefore BC& =\sqrt{122}\\ B{C}^2& =122\end{array}$

Hence, the answer is 122

Example 2: Let $(\alpha ,\beta )$ be the centroid of the triangle formed by the lines $15x-y=82,6x-5y=-4$ and $9x+4y=17$Then $\alpha +2\beta$ and $2\alpha -\beta$ are the roots of the equation [JEE MAINS 2023]

Solution

$\begin{array}{rl}& \text{ Centroid }(\alpha ,\beta )=(\frac{6+1+5}{3},\frac{8-7+2}{3})=(4,1)\\ & \alpha +2\beta =4+2=6\\ & 2\alpha -\beta =8-1=7\end{array}$

Quadratic equation
$\begin{array}{rl}& x^2-(6+7)x+(6\times 7)=0\\ & \Rightarrow x^2-13x+42=0\end{array}$

Hence, the answer is $x^2-13x+42=0$

Example 3: Let R be the focus of the parabola $y^2=20x$ and the line $y=mx+c$ intersects the parabola at two points P and Q. Let the point G(10, 10) be the centroid of the triangle PQR. If $c-m=6$, then $(\mathrm{PQ}{)}^2$ [JEE MAINS 2023]

Solution

$\begin{array}{rl}& y^2=20x,y=mx+c\\ & y^2=20\left(\frac{y-c}{m}\right)\\ & y^2-\frac{20y}{m}+\frac{20c}{m}=0\frac{y_1+y_2+y_3}{3}=10\\ & \frac{20}{m}=30\\ & m=\frac{2}{3}\end{array}$
and $\mathrm{c}-\mathrm{m}=6$

$\begin{array}{rl}& \mathrm{c}=\frac{2}{3}+6\Rightarrow \frac{20}{3}=c\\ & y^2-30\mathrm{y}+\frac{20\times 20/3}{2/3}=0\Rightarrow {\mathrm{y}}^2-30\mathrm{y}+200=0\\ & \mathrm{y}=10,\mathrm{y}=20\\ & \mathrm{y}=20,\mathrm{x}=20P(5,10);(20,20)\mathrm{Q}\\ & \frac{20+5+x}{3}=10\Rightarrow x=5{\mathrm{PQ}}^2={15}^2+{10}^2=225+100=325\end{array}$

Hence, the answer is $325$.

Example 4: In an isosceles triangle ABC, the vertex A is $(6,1)$ and the equation of the base BC is $2x+y=4$.. Let the point B lie on the line $x+3y=7$ If $(\alpha ,\beta )$ is the centroid of $\mathrm{△}\mathrm{ABC}$, then $15(\alpha +\beta )$ is equal to? [JEE MAINS 2022]

Solution: The foot of perpendicular from $\mathrm{A}(6,1)$ to BC

$\frac{x-6}{2}=\frac{y-1}{1}=\frac{-1(6x^2+1-4)}{2^2+1^2}=-\frac{9}{5}$

$\Rightarrow \mathrm{x}=\frac{12}{5},\mathrm{y}=\frac{-4}{5}$

Centroid divide medium in $2:1$

$\frac{\mathrm{A}(6,1)\mathrm{G}(\alpha ,\beta )}{2:1}=\mathrm{D}(\frac{12}{5},\frac{-4}{5})$

$\begin{array}{rl}& \text{ So }\alpha =\frac{6\times 1+\frac{12}{5}\times 2}{2+1},\beta =\frac{1\times 1+2\times (-\frac{4}{5})}{2+1}\\ & \alpha +\beta =\frac{30+24+5-8}{15}=\frac{51}{15}\Rightarrow 15(\alpha +\beta )=51\end{array}$

Hence, the answer is $51$.

Example 5: Let $A(1,0),B(6,2)$ and $C(\frac{3}{2},6)$ be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB, and BPC have equal areas, then the length of the line segment PQ, where Q is the point $(-\frac{7}{6},-\frac{1}{3})$is [JEE MAINS 2020]

Solution:

$A(1,0),B(6,2),C(3/2,6)$

Point $P$ is the centroid of $\mathrm{△}ABC$

$P(\frac{17}{6},\frac{8}{3})$

Distance between PQ is $5$

Hence, the answer is 5.

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