Centroid - Definition, Properties, Theorem and Formulas

In this article, we will cover the concepts of the Centroid of a Triangle. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of seventeen questions have been asked on JEE MAINS (2013 to 2023) from this topic including one in 2014, one in 2020, one in 2021, and three in 2023.

Centroid Definition

The Centroid of a triangle is the point of intersection of the three medians of the triangle. A median is a line joining the mid-points of a side and the opposite vertex of a triangle. The centroid is the center point of the object. A centroid divides any median in the ratio 2:1.

Centroid of Triangle Formula

The point of concurrency of the medians of a triangle is called the centroid of the triangle and always lies in the triangle.

Let us get the coordinates of the centroid in terms of the coordinates of the vertices of the triangle.------

Let $\mathrm{A}\left({x}_1, {y}_1\right)$, ${B}\left({x}_2, \mathrm{y}_2\right)$, and $C\left(x_3, y_3\right)$ 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.

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

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

Note:

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

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

Centroid Theorem

The centroid theorem states that the centroid of the triangle is at $2 / 3$ of the distance from the vertex to the mid-point of the sides.

Centroid of A Right Angle Triangle

The centroid of a right-angle triangle is the point of intersection of three medians, drawn from the vertices of the triangle to the midpoint of the opposite sides.

Centroid of a Square

The point where the diagonals of the square intersect each other is the centroid of the square.

Example

If Origin is the centroid of a triangle ABC, and the coordinates of the other two vertices of the triangle are A (4, –3) and B (–5, 2), then find the coordinates of the third vertex.

Solution

Let point C be $(\alpha, \beta)$

$\begin{aligned} & (0,0)=\left(\frac{\alpha+4-5}{3}, \frac{\beta-3+2}{3}\right) \\ & \Rightarrow \frac{\alpha+4-5}{3}=0 \text { and } \frac{\beta-3+2}{3}=0\end{aligned}$
$\alpha=1$, and $\beta=1$

Coordinates of C are (1, 1)

Important Points

1) The Centroid of the triangle and that of a triangle formed by midpoints of sides are the same.

2) The three medians of a triangle divide the triangle into six equal areas.

3) In a triangle, the line segments connecting the midpoints of sides are parallel to the opposite sides and the four triangles created by these line segments are congruent and so have equal areas.

Solved Examples Based on Centroid

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

Solution

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

put 'B' in BC

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

$p=11$

put 'C' in BC

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

$\begin{aligned} \therefore B C & =\sqrt{122} \\ B C^2 & =122\end{aligned}$

Hence, the answer is 122

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

Solution

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

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

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

Example 3: Let R be the focus of the parabola $y^2=20 x$ and the line $y=m x+c$ 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{aligned}
& \mathrm{y}^2=20 x, y=m x+c \\
& \mathrm{y}^2=20\left(\frac{\mathrm{y}-\mathrm{c}}{\mathrm{m}}\right) \\
& \mathrm{y}^2-\frac{20 \mathrm{y}}{\mathrm{m}}+\frac{20 \mathrm{c}}{\mathrm{m}}=0 \quad \frac{\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3}{3}=10 \\
& \frac{20}{\mathrm{~m}}=30 \\
& \mathrm{~m}=2 / 3
\end{aligned}
$
and $\mathrm{c}-\mathrm{m}=6$

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

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 $2 x+y=4$.. Let the point B lie on the line $x+3 y=7$ If $(\alpha, \beta)$ is the centroid of $\triangle \mathrm{ABC}$, then $15(\alpha+\beta)$ is equal to? [JEE MAINS 2022]

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

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

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

Centroid divide medium in $2: 1$

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

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

Hence, the answer is 51.

Example 5: Let $A(1,0), B(6,2)$ and $C\left(\frac{3}{2}, 6\right)$ 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 $\left(-\frac{7}{6},-\frac{1}{3}\right)$is [JEE MAINS 2020]

Solution:

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

Point P is the centroid of triangle ABC

P(17/6,8/3)

Distance between PQ is 5

Hence, the answer is 5

Summary

The centroid is the centre point of the figure. It is the balance point of the body. It provides us the information about the point where the mass of the body is concentrated. With the help of the section formula, we can find the centroid of the figure.