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Answer:
(a) 2 or 3
(b) -13
Step-by-step explanation:
(a) A, B, and C are collinear if there is some constant 'c' such that ...
B - A = c(C - A)
This will give rise to two equations in 'c' and 'k'.
(k +1, k +4) -(0, 9) = c((2k, k+3) -(0, 9))
(k +1, k -5) = c(2k, k -6)
And the two equations are ...
k +1 = 2ck
k -5 = c(k -6)
From the first, we find ...
c = (k +1)/(2k)
From the second, we find ...
c = (k -5)/(k -6)
Equating these expressions for c gives ...
(k +1)/(2k) = (k -5)/(k -6)
(k +1)(k -6) = 2k(k -5) . . . . . . . multiply by 2k(k-6)
k^2 -5k -6 = 2k^2 -10k
0 = k^2 -5k +6 = (k -3)(k -2)
Solutions that make the factors zero are ...
k = 3 or k = 2
There are two values of k that make the points collinear: 2 and 3.
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(b) AB is parallel to CD when the slopes of the lines between them are the same
slope of AB = (k+4 -9)/(k+1 -0) = (k -5)/(k +1)
slope of CD = ((k +6 -(k +3))/(2k +2 -2k) = 3/2
Equating these slopes and solving for k, we get ...
(k -5)/(k +1) = 3/2
2(k -5) = 3(k +1)
2k -10 = 3k +3
-13 = k
AB is parallel to CD when k = -13.
_____
The attached shows the collinear points on the red and blue lines. The points that make parallel lines are shown on the purple lines. (The purple circles correspond to points D for k=2 and 3, so are irrelevant.) Point D is not labeled with its coordinates, (-24, -7).
