Chapter 8.4, Problem 44WE

To determine

To find: The navigator’s position.

Answer to Problem 44WE

The position of navigator is $\left(\frac{5}{4},-\frac{5}{4}\right)$ .

Explanation of Solution

Given information:

Two radio beacons are located at $\left(-1,0\right)$ and $\left(2,-1\right)$ .

The slope of line joining first and second radio beacon with navigator is -5 and 3 respectively.

Formula used:

Slope of line joining two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$ .

Let the location of navigator be $\left(x,y\right)$

Slope of line joining radio beacon at $\left(-1,0\right)$ and navigator is -5.

Recall that Slope of line joining two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$ .

Slope of line joining radio beacon at $\left(-1,0\right)$ and navigator will be:

  $\begin{array}{l}\frac{y-0}{x-\left(-1\right)}=-5\\ \frac{y}{x+1}=-5\\ y=-5\left(x+1\right)\\ y=-5x+\mathrm{5........}\left(1\right)\end{array}$

Slope of line joining radio beacon at $\left(2,-1\right)$ and navigator is 3.

Recall that Slope of line joining two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$ .

Slope of line joining radio beacon at $\left(2,-1\right)$ and navigator will be:

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

  $y=3x-\mathrm{5.....}\left(2\right)$

Putting value of y from equation 1 to equation 2,

  $\begin{array}{l}-5x+5=3x-5\\ 5+5=3x+5x\\ 10=8x\\ x=\frac{5}{4}\end{array}$

Putting value of x in equation 2,

  $\begin{array}{l}y=3\left(\frac{5}{4}\right)-5\\ y=\frac{15}{4}-5\\ y=\frac{15-20}{4}\\ y=\frac{-5}{4}\end{array}$

The position of navigator is $\left(\frac{5}{4},-\frac{5}{4}\right)$ .