Q. 12 A4.6( 7 Votes )

# The co-ordinates

Answer :

Let the vertices of ΔABC be

A = (x_{1}, y_{1}) and

B = (x_{2}, y_{2}) and

C = (x_{3}, y_{3})

Centroid of a triangle is given by

G =

Let (0, 1), (1, 1) and (1, 0) be midpoints of AB, BC and AC respectively

Point (0, 1) is the midpoint of AB

⇒ (0, 1) =

Equating x and y coordinates

⇒ = 0 and = 1

⇒ x_{1} + x_{2} = 0 … (a) and y_{1} + y_{2} = 2 … (i)

Point (1, 1) is the midpoint of BC

⇒ (1, 1) =

Equating x and y coordinates

⇒ = 1 and = 1

⇒ x_{2} + x_{3} = 2 … (b) and y_{2} + y_{3} = 2 … (ii)

Point (1, 0) is the midpoint of AC

⇒ (1, 0) =

Equating x and y coordinates

⇒ = 1 and = 0

⇒ x_{1} + x_{3} = 2 … (c) and y_{1} + y_{3} = 0 … (iii)

Adding equations (a), (b) and (c) & adding equations (i), (ii) and (iii) we get

⇒ x_{1} + x_{2} + x_{2} + x_{3} + x_{1} + x_{3} = 4 and y_{1} + y_{2} + y_{2} + y_{3} + y_{1} + y_{3} = 4

⇒ 2 (x_{1} + x_{2} + x_{3}) = 4 and 2 (y_{1} + y_{2} + y_{3}) = 4

⇒ x_{1} + x_{2} + x_{3} = 2 and y_{1} + y_{2} + y_{3} = 2

Divide by 3

⇒ = and =

Hence centroid is G =

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