Q. 12 A4.6( 7 Votes )

# The co-ordinates

Let the vertices of ΔABC be

A = (x1, y1) and

B = (x2, y2) and

C = (x3, y3)

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

x1 + x2 = 0 … (a) and y1 + y2 = 2 … (i)

Point (1, 1) is the midpoint of BC

(1, 1) =

Equating x and y coordinates

= 1 and = 1

x2 + x3 = 2 … (b) and y2 + y3 = 2 … (ii)

Point (1, 0) is the midpoint of AC

(1, 0) =

Equating x and y coordinates

= 1 and = 0

x1 + x3 = 2 … (c) and y1 + y3 = 0 … (iii)

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

x1 + x2 + x2 + x3 + x1 + x3 = 4 and y1 + y2 + y2 + y3 + y1 + y3 = 4

2 (x1 + x2 + x3) = 4 and 2 (y1 + y2 + y3) = 4

x1 + x2 + x3 = 2 and y1 + y2 + y3 = 2

Divide by 3

= and =

Hence centroid is G =

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