Example 3 from Algebra and Trigonometry, 4.2 Modeling with Linear Functions

Using a Diagram to Model Distance Between Cities

There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
It might help here to draw a picture of the situation. See Figure 4. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates $(30, 10),$ and Eastborough at $(20, 0) .$

Figure 4

Using this point along with the origin, we can find the slope of the line from Westborough to Agritown.

$m = \frac{10 - 0}{30 - 0} = \frac{1}{3}$

Now we can write an equation to describe the road from Westborough to Agritown.

$W (x) = \frac{1}{3} x$

From this, we can determine the perpendicular road to Eastborough will have slope $m = - 3.$ Because the town of Eastborough is at the point (20, 0), we can find the equation.

$\begin{matrix} E (x) & = & −3 x + b \\ 0 & = & −3 (20) + b & Substitute (20, 0) into the equation . \\ b & = & 60 \\ E (x) & = & −3 x + 60 \end{matrix}$
We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,

$\begin{matrix} \frac{1}{3}x & = & - 3 x + 60 \\ \frac{10}{3} x & = & 60 \\ 10 x & = & 180 \\ x & = & 18 & Substitute this back into W (x) . \\ y & = & W (18) \\ = & \frac{1}{3} (18) \\ = & 6 \end{matrix}$
The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction.

$\begin{matrix} distance & = & \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} \\ = & \sqrt{{(18 - 0)}^{2} + {(6 - 0)}^{2}} \\ \approx & 18.974 \end{matrix}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson?

How did it go?

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solution

Your turn

Solution

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