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COORDINATE GEOMETRY

  • Sum based on Centroid of a Triangle

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Centroid formula

Centroid

Point of concurrence of medians

Consider ΔABC

A

B

C

(x1, y1)

(x2, y2)

(x3, y3)

Q

R

P

Let us draw medians AP, BQ and CR

Segment joining vertex and midpoint of opposite side

∴Coordinates of G are

=

x1

+

x2

3

,

y1

+

y2

3

+

x3

+

y3

G

A (x1, y1)

B (x2, y2)

C (x3, y3)

3 of 4

B

C

A

 

G

(3, 2)

(–2, 1)

(x, y)

5

3

1

3

,

Sol.

By Centroid formula,

=

3

2

3

,

2

+

1

3

x1 = 3,

y1 = 2

x2 = –2,

y2 = 1

A(3, 2), B(–2, 1), G

5

3

1

3

,

Let coordinates of C are (x, y)

x3 = x,

y3 = y

∴Coordinates of G

=

x1

+

x2

3

,

y1

+

y2

3

+

x3

+

y3

+

x

+

y

5

3

1

3

,

=

1

3

,

3

3

+

x

+

y

5

3

1

3

,

Let the co-ordinates of A be (x1, y1)

Let the co-ordinates of B be (x2, y2)

Let the co-ordinates of C be (x3, y3)

5

3

=

1 + x

3

–1

3

=

3 + y

3

Centroid Formula

,

+

x1

x2

+

x3

3

+

y1

y2

3

+

y3

4 of 4

 

B

C

A

G

(3, 2)

(–2, 1)

(x, y)

5

3

1

3

,

Sol.

5

3

=

1 + x

3

–1

3

=

3 + y

3

5

=

1

+

x

5

1

=

x

x

=

4

–1

=

3

+

y

–1

3

=

y

y

=

–4

The coordinates of C are (4, –4)