The variable term in 2x - 4 isa.2b. 3c. 4d. 2x
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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.
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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).
