#### Question

If (−2, 3), (4, −3) and (4, 5) are the mid-points of the sides of a triangle, find the coordinates of its centroid.

#### Solution

Let ΔABC be ant triangle such that P (−2, 3); Q (4,−3) and R (4, 5) are the mid-points of the sides AB, BC, CA respectively.

We have to find the co-ordinates of the centroid of the triangle.

Let the vertices of the triangle be`A(x_1,y_1);B(x_2,y_2);C(x_3,y_3)`

In general to find the mid-point p(x,y) of two points`A(x_1,y_1)`and`B(x_2,y_2)` we use section formula as,

`p(x,y)=((x_1+x_2)/2,(y_1+y_2)/2)`

So, co-ordinates of P,

`(-2,3)=((x_1+x_2)/2,(y_1+y_2)/2)`

Equate the x component on both the sides to get,

`x_1+x_2=-4` .........(1)

Similarly,

`y_1+y_2=6` ..........(2)

Similary, co-ordinates of Q

`(4,-3)=((x_3+x_2)/2,(y_3+y_2)/2)`

Equate the x component on both the sides to get,

`x_3+x_2=8`.........(3)

Similarly,

`y_3+y_2=-6 `..........(4)

Equate the x componet on both the sides to get,

`x_3+x_1=8`..........(5)

Similarly,

`y_3+y_1=10`..........(6)

Add equation (1) (3) and (5) to get,

`2(x_1+x_2+x_3)=12 `

`x_1+x_2+x_3 =6`

Similarly, add equation (2) (4) and (6) to get,

`2(y_1+y_2+y_3)=10`

`y_1+y_2+y_3=5`

We know that the co-ordinates of the centroid G of a triangle whose vertices are

`(x_1,y_1), (x_2,y_2),(x_3,y_3) is `

`G((x_1+x_2+x_3)/3,( y_1+y_2+y_3)/3)`

So, centroid Gof a triangle `triangle ABC `is ,

`G(2,5/3)`