Find the value(s) of $k$ for which the points $(3 k - 1, k - 2), (k, k - 7)$ and $(k - 1, - k - 2)$ are collinear.

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Solution

Given, $A (3 k - 1, k - 2), B (k, k - 7) a n d C (k - 1, - k - 2)$

Two points are said to be collinear if their slopes are equal

$\Rightarrow$ Slope of $A B$ $=$ Slope of $B C$

We have,
$\Rightarrow$ Slope between two points $= (\frac{y_{2} - y_{1}}{x_{2} - x_{1}})$

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

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

$\Rightarrow$ $(5 - 2 k) (1 - 2 k) = 5$

$\Rightarrow$ $5 - 10 k - 2 k + 4 k^{2} = 5$

$\Rightarrow$ $4 k^{2} - 12 k = 0$

$\Rightarrow$ $4 k (k - 3) = 0$

$\Rightarrow$ $k = 0, 3$

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