The points A (k, 2 2k), B ( k + 1, 2k), and C (4 k, 6 2k) are

Question

The points A (k, 2 – 2k), B (– k + 1, 2k), and C (–4 – k, 6 – 2k) are collinear (lie on the same line), then the values of k can be:

A) 2
B) 4
C) 1
D) 5
E) -1

Phoenix19 · Answer

Solve for slope and equate Yb - Ya/Xb - Xa = Yc-Yb/Xc-Xb
 6 - 4K/-5= -2
therefore K=-1

OP1001 · Answer

1) Re-arrange the equation so that the gradients for one pair = the gradient for another pair

e.g.

2-4k/2k-1 = 6-4k/-5

2) use answer choices

Answer choice A (2) = 2 -4(2)/2(2)-1 = 6-4(2)/-5 -> -6/3 = -2/5 (not correct)
Answer choice E (-1) (odd one out with a negative) = 6/-3 = 10/-5 (correct)

bv8562 · Answer

Since all three points lie on same line, each line segment must have same slope i.e line segment, for example, A to B and B to C must have same slope.

Slope of line segment A to B = m1
Slope of line segment B to C = m2

m1 = \frac{2k-(2-2k)}{-k+1-k}

m1 = \frac{4k-2}{1-2k}

m2 = \frac{6k-2k-2k}{-4-k+k-1}

m2 = \frac{6-4k}{-5}

Since, m1=m2, we get:

m1 = \frac{4k-2}{1-2k}  =  m2 = \frac{6-4k}{-5}

upon solving we get two values of k =  \frac{1}{2}  or -1 (E)

bumpbot · Answer

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The points A (k, 2 2k), B ( k + 1, 2k), and C (4 k, 6 2k) are

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