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.

Answer 1

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 equations (1), (3) and (5) to get, 

`2(x_1+x_2+x_3)=12 ` 

`x_1+x_2+x_3 =6` 

Similarly, add equations (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 G of a triangle ABC is `G(2, 5/3)`.