Equation of AB:
To find the equation of line AB, we need to find the slope (m) and the y-intercept (c) of the line.

Slope (m) = (y2 - y1) / (x2 - x1)
Using the coordinates A(10,4) and B(-4,9):
m = (9 - 4) / (-4 - 10) = 5 / -14 = -5/14

Now, we can use the point-slope form of a line equation:
y - y1 = m(x - x1)
Using point A(10,4):
y - 4 = (-5/14)(x - 10)
y - 4 = (-5/14)x + 50/14
y = (-5/14)x + 50/14 + 56/14
y = (-5/14)x + 106/14
y = (-5/14)x + 53/7

So, the equation of line AB is y = (-5/14)x + 53/7.

The Median through A:
A median of a triangle is a line segment that connects a vertex to the midpoint of the opposite side. To find the equation of the median through point A, we need to find the midpoint of side BC.

Midpoint formula:
x = (x1 + x2) / 2
y = (y1 + y2) / 2
Using the coordinates B(-4,9) and C(-2,-1):
Midpoint of BC = ((-4 + -2) / 2, (9 + -1) / 2) = (-3, 4)

Now, we have the coordinates of the midpoint of BC, which is (-3, 4). We can use this point and the coordinates of point A(10,4) to find the equation of the median.

Again, using the point-slope form of a line equation:
y - y1 = m(x - x1)
Using point A(10,4):
y - 4 = (-5/14)(x - 10)
y - 4 = (-5/14)x + 50/14
y = (-5/14)x + 50/14 + 56/14
y = (-5/14)x + 106/14
y = (-5/14)x + 53/7

So, the equation of the median through point A is y = (-5/14)x + 53/7.

The Altitude through B:
The altitude of a triangle is a line segment that is perpendicular to a side of the triangle and passes through the opposite vertex. To find the equation of the altitude through point B, we need to find the slope of the line perpendicular to AB.

The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line.

The slope of AB is -5/14, so the slope of the altitude through B will be 14/5.

Using the point-slope form of a line equation:
y - y1 = m(x - x1)
Using point B(-4,9):
y - 9 = (14/5)(x - (-4))
y - 9 = (14/5)x + (56/5