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
How can the points where the line crosses the farmland be found?
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.
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