Using slope to find radio transmitter signal

Question

A small radio transmitter broadcasts in a 60 mile radius. If you drive along a straight line from a city 79 miles north of the transmitter to a second city 80 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

Answer 1

Hi Nicole:

Here's how to tackle this type of problem.

First, draw a diagram (very important.) Draw it on the usual X-Y axes.

Place the transmitter at the origin, i.e., at (0, 0).

Let the 1st city be due north of the transmitter, at (0, 79).

Let the 2nd city be due east of the transmitter, at (80, 0)

Then, the maximum transmission distance is a circle, center (0, 0) and radius 60.

The signal can be picked up only inside this radius.

So, the question becomes: where does the straight line from (0, 79) to (80, 0) intersect the circle?

The equation of the circle is: x² + y² = 60² [center the origin and radius 60 . . . . . . . . . eqn (1)

The equation of the straight line drive is y = mx + b [where m = slope and b = Y-intercept

m = (y₂ - y₁) / (x₂ - x₁)

= (79 - 0) / (0 - 80)

= -79/80

The straight line's Y-intercept is at (0, 79) => b = 79

∴ Equation of straight line is: y = (-79/80)x + 79

Substitute into eqn (1): x² + ( (-79/80)x + 79)² = 60² which leads to a messy quadratic in x

Use the quadratic formula to find the 2 solutions, that is, the 2 values of x between which the straight line is inside the circle.

I leave you to do the arithmetic!