Chapter 13.7, Problem 39WE

a.

To determine

To proof : the points $C,G\text{ and }H$ are collinear

Answer to Problem 39WE

The points $C,G\text{ and }H$ are collinear $\left(proved\right)$

Explanation of Solution

Given information :

  $\Delta QRS$ With vertices $Q\left(-6,0\right),R\left(12,0\right)\text{ and }S\left(0,12\right)$

Formula used :

  $\text{the slope of the line }=\frac{y-y_1}{x-x_1}$

Calculation :

The point of centroid $G$ is $\left(2,4\right)$

The point of circumcenter $C$ is $\left(3,3\right)$

The point of orthocenter $H$ is $\left(0,6\right)$

The slope of line $CG$

  $\begin{array}{l}=\frac{3-4}{3-2}\\ =\frac{-1}{1}\\ =-1\end{array}$

The slope of line $GH$

  $\begin{array}{l}=\frac{4-6}{2-0}\\ =\frac{-2}{2}\\ =-1\end{array}$

The slope of line $CH$

  $\begin{array}{l}=\frac{3-6}{3-0}\\ =\frac{-3}{3}\\ =-1\end{array}$

Therefore, the slope of $CG,GH\text{ and C}H$ is $-1$

Hence, the points $C,G\text{ and }H$ are collinear $\left(proved\right)$

b.

To determine

To proof : $GH=2GC$

Answer to Problem 39WE

  $GH=2GC\left(proved\right)$

Explanation of Solution

Given information :

  $\Delta QRS$ With vertices $Q\left(-6,0\right),R\left(12,0\right)\text{ and }S\left(0,12\right)$

Calculation :

The distance of $G$ from $H$

  $\begin{array}{l}GH=\sqrt{{\left(2-0\right)}^2+{\left(4-6\right)}^2}\\ GH=\sqrt{{\left(2\right)}^2+{\left(2\right)}^2}\\ GH=\sqrt{4+4}\\ GH=\sqrt{8}\end{array}$

The distance of $G$ from $C$

  $\begin{array}{l}GC=\sqrt{{\left(3-2\right)}^2+{\left(3-4\right)}^2}\\ GC=\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2}\\ GC=\sqrt{1+1}\\ GC=\sqrt{2}\end{array}$

Therefore,

  $\frac{GH}{GC}=\frac{\sqrt{8}}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}}=2$

Therefore, $GH=2GC$

Therefore, the distance of $G$ from $H$ is equal to the twice of the distance of $G$ from $C$

Hence, $GH=2GC\left(proved\right)$