If a vertex of a triangle is (1, 1) and the midpoints of two sides of the triangle through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle is

Concept:

Centroid of a Triangle: The point where the three medians of the triangle intersect.

Let’s consider a triangle. If the three vertices of the triangle are A(x₁, y₁), B(x₂, y₂), C(x₃, y₃).

Centroid of a triangle = \(\left( {\frac{{{{\rm{x}}_1} + {\rm{\;}}{{\rm{x}}_2} + {\rm{\;}}{{\rm{x}}_3}}}{3},\frac{{{{\rm{y}}_1} + {\rm{\;}}{{\rm{y}}_2} + {\rm{\;}}{{\rm{y}}_3}}}{3}} \right)\)

Calculation:

Let’s consider a triangle. If the three vertices of the triangle are A(x₁, y₁), B(x₂, y₂), C(x₃, y₃).

Given: A (x₁, y₁) = A (1, 1)

Mid-Point of AB is given (-1, 2)

\(\frac{{{{\rm{x}}_1} + {\rm{\;}}{{\rm{x}}_2}}}{2} = {\rm{\;}} - 1{\rm{\;and\;}}\frac{{{{\rm{y}}_1} + {\rm{\;}}{{\rm{y}}_2}}}{2} = {\rm{\;}}2{\rm{\;\;}}\) 

⇒ x₂ = -2 – 1 = -3 and y₂ = 4 – 1 = 3

So, B(x₂, y₂) = B (-3, 3)

Mid-Point of AC is given (3, 2)

\(\frac{{{{\rm{x}}_1} + {\rm{\;}}{{\rm{x}}_3}}}{2} = 3{\rm{\;and\;}}\frac{{{{\rm{y}}_1} + {\rm{\;}}{{\rm{y}}_3}}}{2} = {\rm{\;}}2{\rm{\;\;}}\) 

⇒ x₃ = 6 – 1 = 5 and y₃ = 4 – 1 = 3

C(x₃, y₃) = (5, 3)

Centroid of a triangle = \(\left( {\frac{{{{\rm{x}}_1} + {\rm{\;}}{{\rm{x}}_2} + {\rm{\;}}{{\rm{x}}_3}}}{3},\frac{{{{\rm{y}}_1} + {\rm{\;}}{{\rm{y}}_2} + {\rm{\;}}{{\rm{y}}_3}}}{3}} \right) = {\rm{}}\left( {\frac{{1 - 3 + {\rm{\;}}5}}{3},\frac{{1 + {\rm{\;}}3 + {\rm{\;}}3}}{3}} \right) = {\rm{}}\left( {1,\frac{7}{3}} \right)\)