Chapter 13.7, Problem 36WE

a.

To determine

To find : the equation of the three lines that contain the medians

Answer to Problem 36WE

The equation of the three lines $S{M}_1$ , $Q{M}_2$ and $R{M}_3$ are $4x+y=12$ , $x-2y=6$ and $2x+5y=24$

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 :

  $\begin{array}{l}\text{the slope of the line }=\frac{y-y_1}{x-x_1}\text{ and }\\ \text{the equation of the line }y-y_1=mx-x_1\end{array}$

Calculation :

Let $M_1,M_2\text{ and }{M}_3$ be the median of the $QR,RS\text{ and SQ}$ three lines.

To find the medians

  $\begin{array}{l}{M}_1=\left(\frac{12+\left(-6\right)}{2},\frac{0+0}{2}\right)\\ M_1=\left(\frac{12-6}{2},0\right)\\ M_1=\left(\frac{6}{2},0\right)\\ M_1=\left(3,0\right)\\ M_2=\left(\frac{0+12}{2},\frac{12+0}{2}\right)\\ M_2=\left(\frac{12}{2},\frac{12}{2}\right)\\ M_2=\left(6,6\right)\\ \text{and }{M}_3=\left(\frac{-6+0}{2},\frac{0+12}{2}\right)\\ M_3=\left(\frac{-6}{2},\frac{12}{2}\right)\\ M_3=\left(-3,6\right)\end{array}$

The slope of line $S{M}_1$

  $\begin{array}{l}=\frac{0-12}{3-0}\\ =\frac{-12}{3}\\ =-4\end{array}$

The slope of line $Q{M}_2$

  $\begin{array}{l}=\frac{6-0}{6-\left(-6\right)}\\ =\frac{6}{12}\\ =\frac{1}{2}\end{array}$

The slope of line $R{M}_3$

  $\begin{array}{l}=\frac{6-0}{-3-12}\\ =\frac{-6}{12}\\ =-\frac{2}{5}\end{array}$

The equation of the line $S{M}_1$

  $\begin{array}{l}y-0=-4\left(x-3\right)\\ y=-4x+12\\ 4x+y=12\end{array}$

Therefore, the equation of the line $S{M}_1$ is $4x+y=12$

The equation of the line $Q{M}_2$

  $\begin{array}{l}y-0=\frac{1}{2}\left(x-\left(-x\right)\right)\\ y=\frac{1}{2}x+\frac{6}{2}\\ 2y=x+6\end{array}$

Therefore, the equation of the line $Q{M}_2$ is $2y=x+6$

The equation of the line $R{M}_3$

  $\begin{array}{l}y-0=-\frac{2}{5}\left(x-12\right)\\ y=-\frac{2}{5}x+\frac{24}{5}\\ 5y=-2x+24\\ 2x+5y=24\end{array}$

Therefore, the equation of the line $R{M}_3$ is $2x+5y=24$

Hence, the equation of the three lines $S{M}_1$ , $Q{M}_2$ and $R{M}_3$ are $4x+y=12$ , $x-2y=6$ and $2x+5y=24$ .

b.

To determine

To proof : the three medians meet in a point $G$

Answer to Problem 36WE

The three medians meet in a point $G\left(centroid\right)$

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 equation of the line $S{M}_1$ is $4x+y=12\to equation\left(1\right)$

The equation of the line $Q{M}_2$ is $2y=x+6\to equation\left(2\right)$

The equation of the line $R{M}_3$ is $2x+5y=24\to equation\left(3\right)$

From $equation\left(2\right)$

  $y=\frac{x+6}{2}$

Putting the value of $y$ in $equation\left(1\right)$ ,

  $\begin{array}{l}4x+\frac{x+6}{2}=12\\ 8x+x+6=24\\ 9x=24-6\\ 9x=18\\ x=\frac{18}{9}\\ x=2\end{array}$

Putting the value of $x$ in $equation\left(2\right)$ ,

  $\begin{array}{l}2y=2+6\\ 2y=8\\ y=\frac{8}{2}\\ y=4\end{array}$

Putting the value of $x$ in $equation\left(3\right)$ ,

  $\begin{array}{l}2\cdot 2+5y=24\\ 5y=24-4\\ 5y=20\\ y=\frac{20}{5}\\ y=4\end{array}$

Putting the value of $y$ in $equation\left(3\right)$ ,

  $\begin{array}{l}2x+5\cdot 4=24\\ 2x=24-20\\ 2x=4\\ x=\frac{4}{2}\\ x=2\end{array}$

Hence, the three medians meet in a point $G\left(centroid\right)$ .

c.

To determine

To prove : the length $QG$ is $\frac{2}{3}$ of the length of the median from $Q$

Answer to Problem 36WE

The length $QG$ is $\frac{2}{3}$ of the length of the median from $Q\left(proved\right)$

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 :

  $\Delta M_{3}G{M}_2$ And $\Delta QGR$ corresponding angles are equal so, $\Delta M_{3}G{M}_2$ and $\Delta QGR$ are similar.

Therefore, $M_3{M}_2$ is parallel to $QR$

So,

  $M_3{M}_2=\frac{1}{2}QR$

The ratios of corresponding sides are

  $\frac{QG}{G{M}_2}=\frac{RG}{G{M}_3}=\frac{QR}{M_3{M}_2}=\frac{2}{1}\text{ since }{M}_3{M}_2=\frac{1}{2}QR$

Therefore, $QG:G{M}_2=2:1$

The ratio of $QG\text{ and }G{M}_2=2:1$

The length of the side $Q{M}_2=2+1=3$

Therefore, the length of $Q{M}_2=\frac{1}{3}$

And the length of $Q{M}_2=\frac{2}{3}$

Hence, the length $QG$ is $\frac{2}{3}$ of the length of the median from $Q\left(proved\right)$