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PLEASE HELP DUE AT MIDNIGHT

PLEASE HELP DUE AT MIDNIGHT-example-1
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User Surfasb
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(a) To find the equations of medians AD, BE, and CF, we need to first find the coordinates of D, E, and F.

The coordinates of the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) are ((x1 + x2)/2, (y1 + y2)/2).

So, the coordinates of D are:

((Bx + Cx)/2, (By + Cy)/2) = ((-9 + 1)/2, (6 - 4)/2) = (-4, 1)

The coordinates of E are:

((Ax + Cx)/2, (Ay + Cy)/2) = ((5 + 1)/2, (4 - 4)/2) = (3, 0)

The coordinates of F are:

((Ax + Bx)/2, (Ay + By)/2) = ((5 - 9)/2, (4 + 6)/2) = (-2, 5)

Now, we can find the equations of medians AD, BE, and CF.

The equation of the line passing through points (x1, y1) and (x2, y2) is:

y - y1 = ((y2 - y1)/(x2 - x1))(x - x1)

Using point-slope form, we can find the equations of the lines passing through points A, B, and C that are parallel to the medians.

The equation of the line passing through A and D is:

y - 4 = ((1 - 4)/(((-9 + 1)/2) - 5))(x - 5)
y - 4 = 3/4(x - 5)
y = 3/4x - 1/4

The equation of the line passing through B and E is:

y - 6 = ((0 - 6)/(((5 + 1)/2) - (-9)))(x - (-9))
y - 6 = 6/7(x + 9)
y = 6/7x + 120/7

The equation of the line passing through C and F is:

y + 4 = ((5 - (-4))/((-2) - 1))(x - 1)
y + 4 = -3/7(x - 1)
y = -3/7x - 25/7

(b) To show that the medians all pass through the same point, we can find the point of intersection of any two of the medians and then verify that the third median also passes through that point.

Let's find the point of intersection of medians AD and BE. To do this, we need to solve the system of equations:

y = 3/4x - 1/4
y = 6/7x + 120/7

Substituting one equation into the other, we get:

3/4x - 1/4 = 6/7x + 120/7
7(3x/4 - 1/4) = 6x + 720/4 - 7
21x - 28 = 24x + 720 - 28
-3x = -720
x = 240

Substituting x = 240 into either equation, we get y = 179/2.

So, the point of intersection of medians AD and BE is (240, 179/2).

Now, let's check if the third median CF passes through this point.

Substituting x = 240 into the equation of the line passing through C
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User Nsbm
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