If the points $A (k + 1, 2 K), B (3 K, 2 k + 3) a n d C (5 k - 1, 5 k)$ are collinear then find the value of K

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Solution

If points $A (k + 1, 2 k), B (3 k, 2 k + 3)$ & $C (5 k - 1, 5 k)$ are collinear, then

Slope of AB $=$ slope of BC $=$ slope of AC

$\frac{2 k - 2 k - 3}{k + 1 - 3 k} = \frac{2 k + 3 - 5 k}{3 k - 5 k + 1} = \frac{5 k - 2 k}{5 k - 1 - k - 1}$

$\frac{- 3}{- 2 k + 1} = \frac{- 3 k + 3}{1 - 2 k} = \frac{3 k}{4 k - 2}$

On comparing we get
$\frac{- 3}{- 2 k + 1} = \frac{3 k}{4 k - 2}$

$- 12 k + 6 = - 6 k^{2} + 3 k$

$6 k^{2} - 15 k + 6 = 0$

$2 k^{2} - 5 k + 2 = 0$

$k = \frac{+ 5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} = \frac{1}{2}, 2$

Hence value of k can be $(\frac{1}{2} o r 2)$ .

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