Let AD, BE and CF be the medians of the ΔABC.

**The Coordinates of the Centroid**

Consider
a ΔABC whose vertices are *A*(* x*_{1}* *,* y*_{1}),
*B *(* x*_{2}* *,* y*_{2}* *)* *and*
C *(* x*_{3}* *,* y*_{3}* *).

Let *AD*,
*BE* and *CF* be the medians of the Δ*ABC*.

The centroid
*G* divides the median *AD* internally in the ratio 2:1 and therefore
by section formula, the centroid

The centroid G of the triangle with vertices *A*(*x*_{1}
, *y*_{1}), *B* (*x* _{2} , *y*_{2}
) and *C* (*x* _{3} , *y*_{3} ) is

1. Draw ΔABC with vertices A(*x*_{1} , *y*_{1}), B(*x*_{2}, *y*_{2})
and C(*x*_{3}, *y*_{3}) on the graph sheet.

2. Draw medians and locate the centroid of ΔABC

(i) The coordinates of the vertices of ABC where

A(x_{1} , y_{1}) = __20,32__,

B(x_{2} , y_{2}
) = **100,76**

and C(x_{3} , y_{3}
) = *60,144*

(ii) The coordinates of the centroid G = **[****x1+x2+x1]/3 , **[y1+y2+y1]/3

(iii) Use the formula to locate the centroid, whose coordinates are
= **60,84**.

(iv) Mid-point of *AB* is *60,54*.

(v) Find the point which divides the line segment joining (*x*_{3}
, *y*_{3} ) and the mid-point of *AB* internally in the ratio
2:1 is * 60,84*.

**Note**

**• **The medians of a triangle
are concurrent and the point of concurrence, the** **centroid *G*, is one-third
of the distance from the opposite side to the vertex along the median.

**• **The centroid of the triangle
obtained by joining the**
**mid-points of the sides
of a triangle is the same as the centroid of the original triangle.

**• **If (*a*_{1},
*b*_{1}), (*a*_{2}, *b*_{2}) and (*a*_{3},
*b*_{3}) are the mid-points of the sides of a triangle *ABC* then
its centroid *G* is given by

**Example 5.20**

Find the
centroid of the triangle whose veritices are *A*(6, −1),
*B*(8, 3)* *and* C*(10,* *−5).

*Solution*

The centroid
*G*(*x*, *y*) of a triangle whose vertices are (*x*_{1}
, *y*_{1}), (*x*_{2} , *y*_{2} ) and (*x*_{3}
, *y*_{3} ) is given by

We have (*x*_{1}
, *y*_{1} ) = (6, −1); (*x*_{2} , *y*_{2}
) =
(8, 3);

(*x*_{3}
, *y*_{3} ) = (10, −5)

The centroid
of the triangle

**Note**

**• **The Euler line of a triangle
is the line that passes through** **the orthocenter (*H*), centroid (*G*) and the circumcenter
(*S*). *G* divides the line segment in the ratio 2:1
from the orthocenter. That is centroid divides orthocenter and circumcenter internally
in the ratio 2:1 from the Orthocentre.

**• **In an equilateral triangle,
orthocentre, incentre, centroid** **and circumcentre are all the same.

**Example 5.21**

If the centroid
of a triangle is at ( −2,1) and two of its vertices
are (1,
−6)
and (
−5,
2)
, then find the third vertex of the triangle.

*Solution*

Let
the vertices of a triangle be

*A*(1,* *−6),*
B*(−5,
2)* *and* C *(* x*_{3}* *,* y*_{3}* *)

Given the
centroid of a triangle as ( −2,1) we get,

Therefore,
third vertex is (−2,7).

**Thinking
Corner**

(i) Master gave a trianglular plate with vertices A(5, 8), B(2,4),
C(8, 3) and a stick to a student. He wants to balance the plate on the stick. Can
you help the boy to locate that point which can balance the plate.

(ii) Which is the centre of gravity for this triangle? why?

Tags : Formula, Steps, Example Solved Problems | Coordinate Geometry | Maths , 9th Maths : UNIT 5 : Coordinate Geometry

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9th Maths : UNIT 5 : Coordinate Geometry : The Coordinates of the Centroid | Formula, Steps, Example Solved Problems | Coordinate Geometry | Maths

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