Chapter 13.9, Problem 11WE

To determine

To prove: the medians of a triangle intersect in a point called centroid, that is two thirds of the distance from each vertex to the midpoint of the opposite side.

Explanation of Solution

Given information:

Given a triangle with vertices $S(6a,0),R(6b,6c),O(0,0)$ as shown in the figure-

  

Let us name the intersecting point be $D(x,y)$ .

Calculation:

First of all we have to find the coordinate of $D(x,y)$ . In above figure,

  $\begin{array}{l}N=\left(\frac{6b+0}{2},\frac{6c+0}{2}\right)=\left(3b,3c\right)\\ M=\left(\frac{6b+6a}{2},\frac{6c+0}{2}\right)=(3a+3b,3c)\\ J=\left(\frac{6a+0}{2},\frac{0+0}{2}\right)=\left(3a,0\right)\end{array}$

Now,

  $\begin{array}{l}\text{Slope of }\overline{MO}\text{= }\frac{3c-0}{3a+3b-0}=\frac{c}{a+b}\\ \text{Equation of }\overline{MO},\\ \frac{y-0}{x-0}=\frac{c}{a+b}\\ \Rightarrow y=\frac{c}{a+b}x\mathrm{.............................}(i)\end{array}$

Similarly,

  $\begin{array}{l}\text{Slope of }\overline{RJ}\text{=Slope of }\overline{DJ}=\frac{6c-0}{6b-3a}=\frac{2c}{2b-a}\\ \text{Equation of }\overline{DJ},\\ \frac{y-0}{x-3a}=\frac{2c}{2b-a}\\ \Rightarrow y=\frac{2c}{2b-a}(x-3a)\mathrm{.............................}(ii)\end{array}$

Substituting (i) in (ii) we get,

  $\begin{array}{l}\frac{c}{a+b}x=\frac{2c}{2b-a}(x-3a)\\ \Rightarrow x\left(\frac{c}{a+b}-\frac{2c}{2b-a}\right)=-\frac{6ac}{2a-b}\\ \Rightarrow \frac{3acx}{a+b}=6ac\\ \Rightarrow x=2(a+b)\end{array}$

From (i),

  $y=\frac{c}{a+b}\times 2(a+b)=2c$

Thus, $D=(2a+2b,2c)$ .

Next we have to show $\overline{OD}=\frac{2}{3}\overline{OM}$ .

  $\begin{array}{l}\overline{OD}=\sqrt{{(2a+2b-0)}^2+{(2c-0)}^2}=2\sqrt{{(a+b)}^2+c^2}\\ \overline{OM}=\sqrt{{(3a+3b-0)}^2+{(3c-0)}^2}=3\sqrt{{(a+b)}^2+c^2}=\frac{3}{2}\overline{OD}\end{array}$

Thus, $\overline{OD}=\frac{2}{3}\overline{OM}$ is proved.