The coordinates of 4 points are A(0,9), B(k + 1, k + 4), C(2k, k + 3) and D(2k + 2, k + 6). Find the value(s) of k if
(a) the points A, B and C are collinear,
(b) AB is parallel to CD.​

9514 1404 393

  (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).

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.