Chapter 13.9, Problem 14WE

a.

To determine

To prove: points $C,G,H$ are collinear of the given triangle.

Explanation of Solution

Given information :

  $G$be the intersection point of the medians of $\Delta ROS$ , has coordinates $(2a+2b,2c)$ as shown in the figure-

  

Calculation:

First we have to find the coordinate of $H$ , and $C$ . The point $C$ is known as the circumcenter and $H$ is the intersection of altitudes.

Case-1:

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

Case-2:

Next to find the point $H$ , let us consider the figure-

Where $\overline{Rr},\overline{Sp},\overline{Oq}$ are the perpendicular from three vertices $R,S,O$ respectively.

Now,

  $\text{Slope of }\overline{OH}=-\frac{1}{\text{Slope of }\overline{Rs}}=-\frac{1}{\frac{6c-0}{6b-6a}}=\frac{a-b}{c}$

Equation of $\overline{OH}$ ,

  $\frac{y}{x}=\frac{a-b}{c}\Rightarrow y=\frac{a-b}{c}x\mathrm{...............}(ii)$

Again,

  $\text{Slope of }\overline{Hr}=\text{Slope of }\overline{HR}=-\frac{1}{\text{Slope of }\overline{sO}}=-\frac{1}{\frac{0-0}{6a-0}}=\frac{1}{0}$

Equation of $\overline{HR}$ ,

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

Substitute value of $\text{x}$ in (ii) we get, $y=\frac{6b(a-b)}{c}=\frac{6ab-6b^2}{c}$ . Thus $H=\left(6b,\frac{6ab-6b^2}{c}\right)$ .

If points $G,H,C$ are collinear then, $\text{slope of }\overline{CG}\text{= slope of }\overline{GH}$ . Now,

  $\begin{array}{l}\text{Slope of }\overline{CG}=\frac{\frac{3b^2+3c^2-3ab}{c}-2c}{3a-2a-2b}=\frac{3b^2+c^2-3ab}{c(a-2b)}\\ \text{Slope of }\overline{GH}=\frac{2c-\frac{6ab-6b^2}{c}}{2a+2b-6b}=\frac{2c^2-6ab+6b^2}{c(2a-4b)}=\frac{3b^2+c^2-3ab}{c(a-2b)}=\text{Slope of }\overline{CG}\end{array}$

Conclusion: Hence all three points $G,H,C$ are collinear.

b.

To determine

To prove: $\overline{CG}=\frac{1}{3}\overline{CH}$

Explanation of Solution

Given information:

  $G$be the intersection point of the medians of $\Delta ROS$ ,has coordinates $(2a+2b,2c)$

  

Calculation:

  $\begin{array}{l}\overline{CG}\text{=}\sqrt{{(2a+2b-3a)}^2+\left(2c-\frac{3b^2+3c^2-3ab}{c}\right)}=\sqrt{{(2b-a)}^2+{\left(\frac{3ab-3b^2-c^2}{c}\right)}^2}\\ \overline{GH}=\sqrt{{(6b-3a)}^2+\left(\frac{6ab-6b^2}{c}-\frac{3b^2+3c^2-3ab}{c}\right)}=\sqrt{9{(2b-a)}^2+9\left(\frac{3ab-3b^2-c^2}{c}\right)}=3\sqrt{{(2b-a)}^2+{\left(\frac{3ab-3b^2-c^2}{c}\right)}^2}=3\overline{CG}\end{array}$

Conclusion: Hence $\overline{CG}=\frac{1}{3}\overline{CH}$ .