To formulate the system, it's necessary to consider the slope of each line along with at least one point from each line. The two lines will connect each plane's location to their destination airport. It's important to note that the airport's coordinates represent the intersection of these two lines, corresponding to the solution of the system. First, the slope of the line from airplane one to the airport is: m = 2; this can be observed by plotting the two points. From airplane 1's location, the rise is 8 units while the run is 4 units to reach the airport, making the slope 8 divided by 4 = 2. We then insert the slope and point (2,4) into the point-slope form: y - 4 = 2(x - 4), which can be rearranged to standard form 2x - y = 0. For airplane two, the slope to the airport is obtained by observing the vertical decrease of 3 and a horizontal increase of 9 as we move from the airport to airplane 2. We then substitute the slope and the point (15,9) into the point-slope form: y - 9 = -1/3(x - 15), which can be rearranged to the standard form: x + 3y = 42. Consequently, the system of equations is: 2x - y = 0 and x + 3y = 42. Multiplying the first equation by 3 produces a system of: 6x - 3y = 0 and x + 3y = 42. Adding these equations results in the equation 7x = 42. Thus, x = 6, and by substituting this value back into 2x - y = 0, we determine y = 12. Thus, we demonstrate that the airport's coordinates do indeed comprise the solution to our system.
I found that there are All Real Numbers (infinite solutions); what are your choices for answers?
Answer:
Detailed steps:
y – 4 = –1 (x – 2)
y – 7 = –1 (x + 1)
y = –x + 6
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