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A small radio transmitter broadcasts in a 61 mile radius. If you drive along a straight line from a city 68 miles north of the transmitter to a second city 81 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

User MrSpock
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To solve this problem, we need to find the intersection of the circle with a 61-mile radius centered at the transmitter and the straight line connecting the two cities.

First, let's draw a diagram of the situation:

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T (transmitter)

|\

| \

| \

| \

| \

| \

| \

| \

C1 C2

Here, T represents the transmitter, C1 represents the city 68 miles north of the transmitter, and C2 represents the city 81 miles east of the transmitter. We want to find out how much of the straight line from C1 to C2 is within the range of the transmitter.

To solve this problem, we need to use the Pythagorean theorem to find the distance between the transmitter and the straight line connecting C1 and C2. Then we can compare this distance to the radius of the transmitter's range.

Let's call the distance between the transmitter and the straight line "d". We can find d using the formula for the distance between a point and a line:

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d = |(y2-y1)x0 - (x2-x1)y0 + x2y1 - y2x1| / sqrt((y2-y1)^2 + (x2-x1)^2)

where (x1,y1) and (x2,y2) are the coordinates of C1 and C2, and (x0,y0) is the coordinate of the transmitter.

Plugging in the values, we get:

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d = |(81-0)*(-68) - (0-61)*(-68) + 0*0 - 61*81| / sqrt((81-0)^2 + (0-61)^2)

= 3324 / sqrt(6562)

≈ 41.09 miles

Therefore, the portion of the straight line from C1 to C2 that is within the range of the transmitter is the portion of the line that is within 61 miles of the transmitter, which is a circle centered at the transmitter with a radius of 61 miles. To find the length of this portion, we need to find the intersection points of the circle and the line and then calculate the distance between them.

To find the intersection points, we can solve the system of equations:

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(x-0)^2 + (y-0)^2 = 61^2

y = (-61/68)x + 68

Substituting the second equation into the first equation, we get:

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(x-0)^2 + (-61/68)x^2 + 68(-61/68)x + 68^2 = 61^2

Simplifying, we get:

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(1 + (-61/68)^2)x^2 + (68*(-61/68))(x-0) + 68^2 - 61^2 = 0

Solving this quadratic equation, we get:

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x = 12.58 or -79.23

Substituting these values into the equation for the line, we get:

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y = (-61/68)(12.58) + 68 ≈ 5.36

y = (-61/68)(-79.23) + 68 ≈ 148.17

Therefore, the intersection points are approximately (12.58, 5.36) and (-79.23, 148.17). The distance between these points is:

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sqrt((12.58-(-79.23))^2 + (5.36-148.17)^2)

User Ykay
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