Chapter 13.9, Problem 13WE

To determine

To prove: the lines containing the altitudes of $\Delta ROS$ intersect in a point $H\left(6b,\frac{6ab-6b^2}{c}\right)$ .

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:

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 $H=\left(6b,\frac{6ab-6b^2}{c}\right)$ proved.