We are given a scenario where a radio transmitter broadcasts in a 21-mile radius, and we need to calculate how much of the straight-line drive between two cities will be within the transmitter's coverage area. The problem involves applying geometric concepts, specifically using the distance formula in coordinate geometry.

Step 1: Assign coordinates to the cities and transmitter

Let the position of the transmitter be at the origin, $(0, 0)$.

• The first city is 29 miles north of the transmitter. Its coordinates are $(0, 29)$.
• The second city is 26 miles east of the transmitter. Its coordinates are $(26, 0)$.

The two cities are connected by a straight line, and we are asked to find how much of that line lies within the 21-mile radius of the transmitter.

Step 2: Equation of the line connecting the two cities

We can find the equation of the line passing through the points $(0, 29)$ and $(26, 0)$.

The slope of the line is:

$m = \frac{0 - 29}{26 - 0} = \frac{-29}{26}$

Using the point-slope form of the equation of a line:

$y - 29 = \frac{-29}{26}(x - 0)$

Simplifying:

$y = -\frac{29}{26}x + 29$

This is the equation of the line connecting the two cities.

Step 3: Equation of the circle (transmitter's coverage)

The transmitter broadcasts in a circular region with a radius of 21 miles, centered at $(0, 0)$. The equation of this circle is:

$x^2 + y^2 = 21^2 = 441$

Step 4: Find points of intersection between the line and the circle

Substitute the equation of the line $y = -\frac{29}{26}x + 29$ into the equation of the circle $x^2 + y^2 = 441$.

1. Replace $y$ in the circle's equation:

$x^2 + \left(-\frac{29}{26}x + 29\right)^2 = 441$

2. Expand the square:

$x^2 + \left(\frac{841}{676}x^2 - 2 \cdot \frac{29}{26} \cdot 29 \cdot x + 841\right) = 441$

Simplify the terms:

$x^2 + \frac{841}{676}x^2 - \frac{1682}{676}x + 841 = 441$

Multiply the entire equation by 676 to eliminate the denominator:

$676x^2 + 841x^2 - 1682x + 676 \cdot 841 = 676 \cdot 441$

$676x^2 + 841x^2 - 1682x + 568616 = 298716$

Simplify further:

$1517x^2 - 1682x + 269900 = 0$

Step 5: Solve the quadratic equation

Now, we solve the quadratic equation for $x$. We can use the quadratic formula:

$x = \frac{-(-1682) \pm \sqrt{(-1682)^2 - 4(1517)(269900)}}{2(1517)}$

Simplify the discriminant:

$x = \frac{1682 \pm \sqrt{1682^2 - 4 \cdot 1517 \cdot 269900}}{2 \cdot 1517}$

We’ll compute these values to find the points of intersection.It appears that the discriminant is negative, which means the quadratic equation has no real solutions. This implies that the straight line between the two cities does not intersect the boundary of the circle. In other words, the entire drive between the two cities is outside the 21-mile radius of the transmitter.

Thus, you would not pick up any signal during the drive from the city 29 miles north to the city 26 miles east of the transmitter.

Let me know if you'd like more details or have any questions!

Here are some related questions to consider:

1. How would the scenario change if the transmitter had a larger radius?
2. What is the distance between the two cities?
3. How could we calculate the shortest distance between the line and the transmitter?
4. What geometric properties of circles and lines help solve problems like this?
5. How would a curved path between the cities affect the calculation?

Tip: In problems involving distances between geometric shapes, always consider the intersection points using algebraic methods like solving equations or using the distance formula.