The entry and exit points of (2, 3), and (12, 6), and 200 ft. extension of the sprinkler system gives;
(1) The sewer line crosses the farmland at (6.53, 4.36), and (8.48, 4.9)
(2) The longest installable sprinkler system is approximately 172.4 feet
<h3>How can the points where the line crosses the farmland be found?</h3>
1. The slope of the sewer line is found as follows;
- m = (6 - 3)/(12 - 2) = 3/10 = 0.3
The equation of the sewer line can be expressed in point and slope form as follows;
y = 0.3•x - 0.6 + 3
y = 0.3•x + 2.4
The equation of the circumference of the sprinkler can be expressed as follows;
Therefore;
(x - 8)² + (0.3•x + 2.4 - 3)² = 2²
Solving gives;
x= 6.53, or x = 8.48
y = 0.3×6.53 + 2.4 = 4.36
y = 0.3×8.48 + 2.4 = 4.9
Therefore;
- The sewer line crosses the farmland at (6.53, 4.36), and (8.48, 4.9)
2. When the farmland does not cross the sewer line, we have;
sewer line is tangent to circumference of farmland
Slope of radial line from center of the land is therefore;
m1 = -1/0.3
Equation of the radial line to the point the sewer line is tangent to the circumference is therefore;
y - 3 = (-1/0.3)×(x - 8)
Which gives;
y = (-1/0.3)×(x - 8) + 3
The x-coordinate is therefore;
0.3•x + 2.4 = (-1/0.3)×(x - 8) + 3
- y = 0.3 × 7.5 + 2.4 ≈ 4.65
The longest sprinkler system is therefore;
d = √((7.5 - 8)² + (4.65 - 3)²) ≈ 1.724
Which gives;
- The longest sprinkler system is 1.724 × 100 ft. ≈ 172.4 ft.
Learn more about the equation of a circle here:
brainly.com/question/10368742
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