The straight road crosses the boundaries of the range of the ambulance service at two points
The length of the road that lies within the range of the ambulance service is approximately 37.95
Reason:
The function that represents the ambulance circular service area is x² + y² = 400
The equation that represents the straight road, y = 3·x + 20
Required:
Length of the road that lies within the range of he ambulance
Solution:
The length of the road is given by the distance between the intersection
points of the road and the ambulance service region as follows;
At the intercepts, we have;
The x, and y-values of the line and the circle are equal, therefore, we have;
At the intercept, y = 3·x + 20, which gives;
x² + y² = 400
x² + (3·x + 20)² = 400
x² + 9·x² + 120·x + 400 = 400
10·x² + 120·x + 400 - 400 = 0
10·x² + 120·x = 0
x² + 12·x = 0
x² + 12·x = x·(x + 12) = 0
When x = 0, y = 3×0 + 20 = 20
When x = -12, y = 3×(-12) + 20 = -16
The intersection points of the line and the circle are;
The length of the segment AB, d, is given as follows;
The length of the road that lies within the range of the ambulance service, d, given to the nearest hundredth = 12·√(10) ≈ 37.95
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