Chapter 13.3, Problem 14PSC

To determine

To find: The length of the altitude from $\text{A}\text{to}\overline{\text{BC}}$ .

Answer to Problem 14PSC

The point of intersection is $\left(2,-2\right)$ .

Explanation of Solution

Given:

In $\Delta \text{ABC}$ , given coordinates are:

  $\text{A}=\left(5,-1\right),\text{B}=\left(1,1\right),\text{C}=\left(5,-11\right)$ .

Calculation:

The slope of $\overline{\text{BC}}$ is calculated as:

  $\begin{array}{c}\text{Slope }\overline{\text{BC}}=\frac{1-\left(-11\right)}{1-5}\\ =\frac{12}{-4}\\ =-3\end{array}$

The equation of $\overline{\text{BC}}$ is given as:

  $\begin{array}{c}y=-3x+b\\ 1=-3\left(1\right)+b\because \text{line passes through }\left(1,1\right)\\ b=4\\ \text{so},y=-3x+4\left(1\right)\end{array}$

The slope of altitude from $\text{A}\text{to}\overline{\text{BC}}$ is given as:

  $\text{slope}=\frac{1}{3}\because \text{slope of altitude}\times \text{slope of }\overline{\text{BC}}=-1$

The equation of altitude from $\text{A}\text{to}\overline{\text{BC}}$ is given as:

  $\begin{array}{c}y=\frac{1}{3}x+b\\ -1=\frac{1}{3}\left(5\right)+b\because \text{line passes through }\left(5,-1\right)\\ b=\frac{-8}{3}\\ \text{so},y=\frac{1}{3}x-\frac{8}{3}\left(2\right)\end{array}$

The point of intersection of equation is calculated as:

  $\begin{array}{c}-3x+4=\frac{1}{3}x-\frac{8}{3}\left[\text{put both equations equivallent}\right]\\ x-8=-9x+12\\ 10x=20\\ x=2\end{array}$

Further simplified as:

  $\begin{array}{c}y=-3\left(2\right)+4\left[\text{substitute }x=2\text{ in equation }\left(4\right)\right]\\ =-2\end{array}$

Therefore the point of intersection $\left(1\right)\&\left(2\right)$ is $\left(2,-2\right)$ .