# Trigonometry

1. A, B and C are collinear, and B is between A and C. The ratio of AB to AC is 3:4. If A is at (-8,1) and B is at (-2,-2), what are the coordinates of C?

2. The segment joining (1,-3) and (4,-6) is extended a distance equal to one-sixth of its own length. Find the terminal point.

I solved these two problems yet neither of them is correct.

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1. AB:AC = 3:4 means that B is 3/4 of the way from A to C.
Think about it. Start at A. When you get to B, AB is 3/4 of the whole distance AC.
B = A + k (C-A)
when k=0, you are at A
when k=1, you are at C
Bx = Ax + 3/4 (Cx - Ax) = -8 + 3/4 (-2+8) = -8 + 3/4 (6) = -7/2
That is, the distance from -8 to -2 is 6. 3/4 of 6 = 9/2
By = 1 + 3/4 (-2-1) = 1 + 3/4 (-3) = -5/4

For #2, If we let A = (1,-3) and B = (4,-6), we want C such that BC = 1/6 AB
That is, AC:AB = 7/6
Cx = Ax + 7/6 (Bx - Ax) = 1 + 7/6 (4-1) = 9/2
Cy = -3 + 7/6 (-6+3) = -3 + 7/6 (-3) = -13/2
So, C = (9/2, -13/2)

You really should get out some graph paper and plot the points, to make sure your answer satisfies the conditions. It may be easier to see geometrically.

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