Answer to Question #130545 in Geometry for Delilah

Question

A triangle has vertices A(-1, 2), B(4, -4), and C(1, 2). Show, using analytic geometry, that the centroid divides each median in the ratio 2:1

Answer 1

The Coordinates of the Centroid O(x,y) of a triangle with Vertices A(-1,2), B(4,-4) and C(1,2).

D, E, and F are Mid-point of AB, BC, CA

So,

$D=\{(4-1)\div 2, (2-4)\div2 \}=(3/2,-1)$

$E=\{(4+1)\div 2, (-4+2)\div2 \}=(5/2,-1)$

$F=\{(-1+1)\div 2, (2+2)\div2 \}=(0,2)$

The Centroid of Triangle $O=\{(x_{1}+x_{2}+x_{3})\div3, (y_{1}+y_{2}+y_{3})\div3\}$

$O=\{(-1+4+1)\div3, (2-4+2)\div3\}$

$O=(4/3,0)$

let the required ratio k:1

O(4/3,0) divides the interval CD in the ratio k:1

x=4/3 then

x={(1)(1)+(3/2)(k) }/(k+1)

4/3=(1+3k/2)/(k+1)

4k+4=3(2+3k)/2

8k+8=6+9k

K=2 So ratio is 2:1

Now

y=0 then y=(2--1K)/(k+1)

0=2-k

k=2 So ratio is 2:1

O(4/3,0) divides the interval BF in the ratio k:1

x=4/3 then

x={(1)(4)+(k)(0)}/(k+1)

4/3=4/k+1

4K+4=12

K=2 so ratio is 2:1

Now

y=0 then

y=(-4+2k)/(k+1)

0=-4+2k

k=2 So ratio is 2:1

O(4/3,0) divides the interval AE in ratio k:1

x=4/3 then

x={(1)(-1)+(5/2)(k)}/(k+1)

4/3=(-1+5k/2)/(k+1)

4k+4=3(-2+5k)/2

8k+8=-6+15k

7k=14

k=2 so ratio is 2:1

Now

y=0 then

y=(2-1k)/(k+1)

0=2-1k

k=2 So ratio is 2:1

Hence

The Centroid divides each Median in the Ratio 2:1