Question

Vertex of a triangle are (4, 6), (2,-2) and (0, 2), then co-ordinates of its centroid must be​

Answer 1

$\setlength{\unitlength}{7.5 mm}\begin{picture}(10, 6)\multiput(2, -2)(-2, 4){2}{\circle*{0.1}}\multiput(4, 6)(-2, -4){2}{\circle*{0.1}}\put(2, -2){\line(1, 4){2}}\put(2, -2){\line(-1, 2){2}}\put(0, 2){\line(1, 1){4}}\put(0, 2){\line(1, 0){3}}\put(2, 2){\line(1, 2){2}}\put(2, 2){\line(0, -1){4}}\put(4.2, 6){ (x_1,\ y_1) }\put(2.2, -2){ (x_2,\ y_2) }\put(-1.8, 2){ (x_3,\ y_3) }\end{picture}$

Given picture is a triangle of vertices with coordinates (x₁, y₁), (x₂, y₂) and (x₃, y₃). Three medians are drawn in the triangle.

The point at which the three medians of a triangle intersect is the centroid of the triangle. We know that the centroid divides each median in the ratio 2:1 measured from the vertex.

Consider the side of the triangle joining points (x₁, y₁) and (x₂, y₂). The point where the median from the opposite vertex of this side meets is the midpoint of the side. We can easily find the coordinates of this midpoint, and it is,

$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

Consider this median, which joins (x₃, y₃) and  $\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right).$

$\setlength{\unitlength}{7.5 mm}\begin{picture}(10, 6)\put(0, 0){\line(1, 0){9}}\multiput(0, 0)(9, 0){2}{\circle*{0.1}}\put(6, 0){\circle*{0.1}}\put(-1.8, 0){ (x_3,\ y_3) }\put(9.2, 0){ \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right) }\put(3, 0.2){2}\put(7.5, 0.2){1}\end{picture}$

By using section formula we can find the coordinates of the centroid.

$\begin{aligned}&\left(\dfrac{2(\frac{x_1+x_2}{2})+x_3}{2+1}, \dfrac{2(\frac{y_1+y_2}{2})+y_3}{2+1}\right)\\ \\ \Longrightarrow\ \ &\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)\end{alinged}$

Hence we get a general formula for finding the coordinates of the centroid.

Now, let,

→  (x₁, y₁) = (4, 6)     ⇒     x₁ = 4   &   y₁ = 6

→  (x₂, y₂) = (2, -2)     ⇒     x₂ = 2   &   y₂ = -2

→  (x₃, y₃) = (0, 2)     ⇒     x₃ = 0   &   y₃ = 2

So, the coordinates of the centroid must be,

$\begin{aligned}&\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)\\ \\ \Longrightarrow\ \ &\left(\frac{4+2+0}{3},\ \frac{6-2+2}{3}\right)\\ \\ \Longrightarrow\ \ &\left(\frac{6}{3},\ \frac{6}{3}\right)\\ \\ \Longrightarrow\ \ &\large \ \textbf{(2, 2)}\end{aligned}$

Hence, (2, 2) is the answer.