Chapter 13.7, Problem 37WE

a.

To determine

To find : the equation of the three perpendicular bisectors of the sides of $\Delta QRS$

Answer to Problem 37WE

The equation of the three lines $SR$ , $RQ$ and $QS$ are $y=x$ , $x=3$ and $x+2y=9$

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 slope of the lines $SR,RQ\text{ and QS}$ .

The slope of line $SR$

  $\begin{array}{l}=\frac{12-0}{0-12}\\ =\frac{12}{-12}\\ =-1\end{array}$

Therefore,

  $\begin{array}{l}{m}_1=\frac{-1}{\text{the slope of }SR}\\ m_1=\frac{-1}{-1}\\ m_1=1\end{array}$

The slope of line $RQ$

  $\begin{array}{l}=\frac{0-0}{12-\left(-6\right)}\\ =\frac{0}{18}\\ =0\end{array}$

Therefore,

  $\begin{array}{l}{m}_2=\frac{-1}{\text{the slope of }RQ}\\ m_2=\frac{-1}{0}\\ m_2=-\infty \end{array}$

The slope of line $QS$

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

Therefore,

  $\begin{array}{l}{m}_3=\frac{-1}{\text{the slope of }QS}\\ m_3=\frac{-1}{2}\end{array}$

The equation of the line $SR$

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

Therefore, the equation of the line $SR$ is $y=x$

The equation of the line $RQ$

  $\begin{array}{l}y-0=-\infty \left(x-3\right)\\ \frac{y}{-\infty }=x-3\\ x-3=0\\ x=3\end{array}$

Therefore, the equation of the line $RQ$ is $x=3$

The equation of the line $QS$

  $\begin{array}{l}y-6=-\frac{1}{2}\left(x-\left(-3\right)\right)\\ y=-\frac{1}{2}x-\frac{3}{2}+6\\ y=-\frac{1}{2}x+\frac{9}{2}\\ 2y=-x+9\\ x+2y=9\end{array}$

Therefore, the equation of the line $QS$ is $x+2y=9$

Hence, the equation of the three lines $SR$ , $RQ$ and $QS$ are $y=x$ , $x=3$ and $x+2y=9$ .

b.

To determine

To proof : the three perpendicular bisectors meet in a point $C$

Answer to Problem 37WE

The three perpendicular bisectors meet in a point $C\left(proved\right)$

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 equation of the line $SR$ is $y=x\to equation\left(1\right)$

The equation of the line $RQ$ is $x=3\to equation\left(2\right)$

The equation of the line $QS$ is $x+2y=9\to equation\left(3\right)$

From $equation\left(2\right)$

  $x=3$

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

  $y=3$

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

  $\begin{array}{l}3+2y=9\\ 2y=9-3\\ y=\frac{6}{2}\\ y=3\end{array}$

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

  $\begin{array}{l}x+2\cdot 3=9\\ x=9-6\\ x=3\end{array}$

Hence, the three perpendicular bisectors meet in a point $C\left(proved\right)$

c.

To determine

To proof : $C$ is equidistance from $Q,R\text{ and }S$ by using the distance formula

Answer to Problem 37WE

The circumcenter $C$ is equidistance from $Q,R\text{ and }S\left(proved\right)$

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 :

From second part

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

The distance of $C$ from $Q$

  $\begin{array}{l}CQ=\sqrt{{\left(3-\left(-6\right)\right)}^2+{\left(3-0\right)}^2}\\ CQ=\sqrt{{\left(9\right)}^2+{\left(3\right)}^2}\\ CQ=\sqrt{81+9}\\ CQ=\sqrt{90}\end{array}$

The distance of $C$ from $R$

  $\begin{array}{l}CR=\sqrt{{\left(3-12\right)}^2+{\left(3-0\right)}^2}\\ CR=\sqrt{{\left(9\right)}^2+{\left(3\right)}^2}\\ CR=\sqrt{81+9}\\ CR=\sqrt{90}\end{array}$

The distance of $C$ from $S$

  $\begin{array}{l}CS=\sqrt{{\left(3-0\right)}^2+{\left(3-12\right)}^2}\\ CS=\sqrt{{\left(3\right)}^2+{\left(-9\right)}^2}\\ CS=\sqrt{9+81}\\ CS=\sqrt{90}\end{array}$

Therefore, the distance of $CQ$ , $CR$ and $CS$ is $\sqrt{90}$

Hence, the circumcenter $C$ is equidistance from $Q,R\text{ and }S\left(proved\right)$

d.

To determine

To find : the equation of the circle that can be circumscribed about $\Delta QRS$

Answer to Problem 37WE

The equation of the circle is $x^2+y^2-6x-6y=72$

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 center of is $C$ , the point of $C$ is $\left(3,3\right)$

Let radius $=r$

The radius of the circle is equal to $CQ,CR\text{ and }CS$

Therefore, $CQ=CR=CS=\sqrt{90}=r$

The equation of the circle is

  $\begin{array}{l}{\left(x-3\right)}^2+{\left(y-3\right)}^2={\left(\sqrt{90}\right)}^2\\ x^2-6x+9+y^2-6y+9=90\\ x^2+y^2-6x-6y=90-18\\ x^2+y^2-6x-6y=72\end{array}$

Hence, the equation of the circle is $x^2+y^2-6x-6y=72$ .