Chapter 13.9, Problem 12WE

To determine

To prove: the perpendicular bisectors of the three sides of $\Delta ROS$ meet in a point $C=\left(3a,\frac{3b^2+3c^2-3ab}{c}\right)$ called circumcenter of the triangle.

Explanation of Solution

Given information:

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

  

Calculation:

To find the circumcenter $C$ consider the following figure where $p,q,r$ are the mid points of $RO,OS$ , and $SR$ respectively. Therefore,

  $\begin{array}{l}p=\left(\frac{6b+0}{2},\frac{6c+0}{2}\right)=\left(3b,3c\right)\\ q=\left(\frac{6b+6a}{2},\frac{6c+0}{2}\right)=(3a+3b,3c)\\ r=\left(\frac{6a+0}{2},\frac{0+0}{2}\right)=\left(3a,0\right)\end{array}$

  

Now,

  $\text{Slope of }\overline{pC}\text{ = }\frac{-1}{\text{Slope of }\overline{RO}}=\frac{-1}{\frac{6c-0}{6b-0}}=-\frac{b}{c}$

Equation of $\overline{pC}$ ,

  $\begin{array}{l}\frac{y-3c}{x-3b}=-\frac{b}{c}\\ \Rightarrow y-3c=-\frac{b}{c}(x-3b)\\ \Rightarrow y=-\frac{b}{c}x+\frac{3b^2}{c}+3c\\ \Rightarrow y=-\frac{b}{c}x+\frac{3(b^2+c^2)}{c}\mathrm{.....................}(i)\end{array}$

Similarly equation of $\overline{Cr}$ ,

  $\begin{array}{l}\frac{y-0}{x-3a}=-\frac{1}{\frac{0-0}{6a-0}}\\ \Rightarrow \frac{y}{x-3a}=\frac{1}{0}\\ \Rightarrow x-3a=0\text{ (Apply method of cross multiplication)}\\ \Rightarrow x=3a\end{array}$

Substitute the value of $x$ in (i) we get,

  $y=-\frac{3ab}{c}+\frac{3(b^2+c^2)}{c}=\frac{3b^2+3c^2-3ab}{c}$

Hence, $C=\left(3a,\frac{3b^2+3c^2-3ab}{c}\right)$ is proved.