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)`
