Chapter 13.7, Problem 38WE

a.

To determine

To find : the equation of the three lines that contain that altitudes of $\Delta QRS$

Answer to Problem 38WE

The equation of the three lines $S{P}_1$ , $Q{P}_2$ and $R{P}_3$ are $x=0$ , $y=x+6$ and $x+2y=12$

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 $P_1,P_2\text{ and }{P}_3$ be the altitudes of lines

Let $m_1,m_2\text{ and }{m}_3$ be the slope of the lines $S{P}_1,Q{P}_2\text{ and }R{P}_3$

The slope of $S{P}_1$

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

Therefore,

  $\begin{array}{l}{m}_1=\frac{-1}{\text{the slope of }S{P}_1}\\ m_1=\frac{-1}{0}\\ m_1=-\infty \end{array}$

The slope of $Q{P}_2$

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

Therefore,

  $\begin{array}{l}{m}_2=\frac{-1}{\text{the slope of }Q{P}_2}\\ m_2=\frac{-1}{-1}\\ m_2=1\end{array}$

The slope of line $R{P}_3$

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

Therefore,

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

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

  $\begin{array}{l}y-12=-\infty \left(x-0\right)\\ \frac{y-12}{-\infty }=x-0\\ 0=x-0\\ x=0\end{array}$

Therefore, the equation of the line $S{P}_1$ is $x=0$

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

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

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

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

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

Therefore, the equation of the line $R{P}_3$ is $x+2y=12$

Hence, the equation of the three lines $S{P}_1$ , $Q{P}_2$ and $R{P}_3$ are $x=0$ , $y=x+6$ and $x+2y=12$ .

b.

To determine

To proof : the three altitudes meet in a point $H$

Answer to Problem 38WE

The three altitudes meet in a point $H\left(proved\right)$

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 $S{P}_1$ is $x=0\to equation\left(1\right)$

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

The equation of the line $R{P}_3$ is $x+2y=12\to equation\left(3\right)$

From $equation\left(1\right)$

  $x=0$

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

  $\begin{array}{l}y=0+6\\ y=6\end{array}$

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

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

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

  $\begin{array}{l}x+2\cdot 6=12\\ x=12-12\\ x=0\end{array}$

Hence, the three altitudes meet in a point $H\left(proved\right)$