Question

find the value of k for the points [3k-1,k-2] [k,k-7] [k-1,-k-2] are collinear in a line

Answer 1

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Answer 2

Answer:

The value of k is either 0 or 3.

Step-by-step explanation:

The given points are A[3k-1,k-2], B[k,k-7], C[k-1,-k-2].

If points A, B and C are collinear, then slope of AB is equal to slope of BC.

Slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

$m_{AB}=m_{BC}$

$\frac{k-7-(k-2)}{k-(3k-1)}=\frac{-k-2-(k-7)}{k-1-k}$

$\frac{k-7-k+2}{k-3k+1)}=\frac{-k-2-k+7}{k-1-k}$

$\frac{-5}{-2k+1}=\frac{-2k+5}{-1}$

$5=(-2k+1)(-2k+5)$

$5=4k^2-12k+5$

$0=4k(k-3)$

Equate each factor equal to zero.

$k=0,3$

Therefore the value of k is either 0 or 3.