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A triangle ABC has sides BC, CA, and AB along the lines x - 2y + 5 = 0, x + y + 2 = 0, and 8x - y - 20 = 0 respectively. If AD is the median from A to BC, find the equation of the line passing through (2, -1) and parallel to AD.

1. 4x - 3y - 11 = 0
2. 13x - 4y - 30 = 0
3. 4x + 13y + 5 = 0
4. 13x + 4y - 22 = 0

Correct answer: 13x + 4y - 22 = 0

Solution

Vertex A is the intersection of lines CA and AB: solving x+y+2=0 and 8x-y-20=0 gives A=(2,-4). Vertex B is the intersection of BC and AB: solving x-2y+5=0 and 8x-y-20=0 gives B=(3,4). Vertex C is the intersection of BC and CA: solving x-2y+5=0 and x+y+2=0 gives C=(-3,1). Midpoint D of BC = ((3-3)/2, (4+1)/2) = (0, 5/2). Slope of AD = (5/2-(-4))/(0-2) = (13/2)/(-2) = -13/4. Line through (2,-1) with slope -13/4: 4(y+1) = -13(x-2) => 13x+4y-22=0.

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