Chapter 11, Problem 52RE

To determine

To find: The x and y coordinates of ship

Answer to Problem 52RE

  $\left(40,41,5\right)$ are the coordinates of ship exhibiting the location of the ship.

Explanation of Solution

Given information:

Distance between A and B $=300\text{ mi}$ . The distance of ship from short line i.e. y − axis is $40\text{ mi}$ .

Calculation:

Distance between $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ is

  $AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Since A & B are on y − axis and are at a distance of $300$ miles, the coordinate of A & B are $\left(0,150\right)$ and $\left(0,-150\right)$ respectively. Here x − axis is half way between them.

Since ship $40\text{ miles}$ away from y − axis i.e. short line, the x − coordinate will be $40$ .

Let y coordinates be y

  $\therefore$ coordinate of ship $S\left(40,y\right)$

It is given that, ship is $80\text{ miles}$ closes to A then B.

  

That means $AS+80=BS$

  $\begin{array}{l}\text{AS}=\text{Distance between A \& B}\\ \text{=}\sqrt{{\left(0-40\right)}^2+{\left(150-y\right)}^2}\\ AS=\sqrt{{\left(40\right)}^2+{\left(150-y\right)}^2}=\sqrt{1600+{\left(150-y\right)}^2}\\ BS=\sqrt{{\left(40-0\right)}^2+{\left(y-\left(-150\right)\right)}^2}\\ BS=\sqrt{1600+{\left(y+150\right)}^2}\end{array}$

We have, $AS+80=BS$

  $\begin{array}{l}BS-AS=80\\ \sqrt{1600+{\left(y+150\right)}^2}-\sqrt{1600+{\left(y-150\right)}^2}=80\\ \sqrt{y^2+300y+24100}-\sqrt{y^2-300y+24100}=80\end{array}$

Interpretation :

If we draw the graph for the above equation it would be $y=41\cdot 5$

At $x=40$ , the y − coordinate will be $\left(40,41\cdot 5\right)$