Final answer:
The equation of the private plane's flight path, which is perpendicular to the jetliner's path of y = (1/2)x - 3, is y = -2x - 2.
Step-by-step explanation:
The student is asking for the equation of the flight path of a private plane which flies perpendicular to the flight path of a jetliner. The jetliner's path is given by the equation y = (1/2)x - 3. To find the equation of the private plane's path that is perpendicular, we first need to determine the negative reciprocal of the jetliner's slope. Since the slope (m) of the jetliner's path is 1/2, the slope of the private plane's path will be -2 (negative reciprocal of 1/2).
Knowing the slope of the private plane and the point (-2, 2) through which the plane's path passes, we can use the point-slope formula, y - y1 = m(x - x1), where (x1, y1) is a point on the line (in this case, the point over the park), and m is the slope of the line. Substituting (-2, 2) for (x1, y1) and -2 for m, we get:
y - 2 = -2(x + 2).
Distributing on the right side, we have:
y - 2 = -2x - 4.
Adding 2 to both sides to put it in slope-intercept form, we obtain:
y = -2x - 2.
So, the equation of the private plane's flight path is y = -2x - 2, which is perpendicular to the jetliner's path.