For what values of $k$ are points $(8, 1) (3, - 2 k)$ and $(k, - 5)$ collinear.

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Solution

Given $(8, 1) (3, - 2 k)$ & $(k, - 5)$ are collinear then slopes are equal

$\Rightarrow \frac{- 2 k - 1}{3 - 8} = \frac{- 5 + 2 k}{k - 3}$

$\Rightarrow (2 k + 1) (k - 3) = 5 (2 k - 5) \Rightarrow 2 k^{2} - 6 k + k - 3 = 10 k - 25$

$\Rightarrow 2 k^{2} - 15 k + 22 = 0$

$\Rightarrow (k - 2) (2 k - 11) = 0$

Hence $[k = 2] o r [k = \frac{11}{2}]$

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Similar questions

A (8, 1)

B (3, - 2 k)

C (k, - 5)

For what value of k are the points $(k, 2 - 2 k), (- k + 1, 2 k), (- 4 - k, 6 - 2 k)$ collinear?

k

(k, 2 - 2 k)

(1 - k, 2 k)

(- 4 - k, 6 - 2 k)

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