Question

Find the centroid of the triangle whose vertices are (2, 3), (4, 5) and (3, 4)?

Answer 1

$x = \frac{2 + 4 + 3}{ 3} \\ = x = 3 \\ x = \frac{3 + 5 + 4}{3} \\ = x = 4$

Answer 2

Question:

Find the centroid of the triangle whose vertices are (2, 3), (4, 5) and (3, 4)?.

Formula Used:

$\sf Centroid \ (x , y) = \Bigg( \dfrac{x_1 + x_2 + x_3}{3} , \dfrac{y_1 + y_2 + y_3}{3} \Bigg)$

Given:

Co-ordinates of the Δgle

(Let us take the vertices to be A, B & C)

A(2, 3)

B(4, 5)

C(3, 4)

Solution:

x₁ → 2

x₂ → 4

x₃ → 3

y₁ → 3

y₂ → 5

y₃ → 4

Let us name the Centroid to be 'D', and let its co-ordinates be D(x, y).

$\Longmaptsto \sf D(x, y) = \Bigg( \dfrac{x_1 + x_2 + x_3}{3} , \dfrac{y_1 + y_2 + y_3}{3} \Bigg)\\ \\ \\ \\\Longmaptsto \sf D(x, y) = \Bigg( \dfrac{2 + 4 + 3}{3} , \ \dfrac{3 + 5 + 4}{3} \Bigg)\\ \\ \\ \\\Longmaptsto \sf D(x, y) = \Bigg( \dfrac{9}{3} , \ \dfrac{12}{3} \Bigg)\\ \\ \\ \\\Longmaptsto \sf D(x, y) = \bigg(3 , 4\bigg)$

Final Answer: D(3, 4)