(a) Let D, E, and F be the midpoints of BC, AC, and AB, respectively. Then the coordinates of D, E, and F are:
D = (-5, 3) E = (3, -1) F = (-1, 0)
The equation of the median AD is:
y - 4 = \frac{3 - 4}{-5 - 5} (x - 5) y - 4 = -\frac{1}{5} (x - 5) 5y - 20 = -x + 5 x + 5y = 25
The equation of the median BE is:
y - (-1) = \frac{0 - (-1)}{3 - 1} (x - 3) y + 1 = \frac{1}{2} (x - 3) 2y + 2 = x - 3 x - 2y = 5
The equation of the median CF is:
y - 0 = \frac{-4 - 4}{1 - (-9)} (x - 1) y = \frac{-8}{10} (x - 1) y = -\frac{4}{5} (x - 1) 5y = -4x + 5 4x - 5y = -5
(b) Let G be the intersection of two of the medians. We want to show that G lies on the third median.
Without loss of generality, let's assume that G is the intersection of AD and BE. Then the coordinates of G satisfy both the equations of AD and BE.
Substituting the equation of AD into the equation of BE, we get:
x - 2y = 5 4x - 5y = -5
Solving for x and y, we get x=0 and y=1. This means that G is the point (0,1), which is on the third median CF.
Therefore, all three medians pass through the same point.