If the points (k,2-2k), (1-k, 2k) and (-k-4, 6-2k) are collinear, then a possible values of k is .

Open in App

Solution

Points A(k,2-2k), B (1-k, 2k) and C(-k-4, 6-2k) are collinear.
Then, slope of line AB = slope of line BC
$\Rightarrow \frac{2 k - (2 - 2 k)}{1 - k - k} = \frac{6 - 2 k - 2 k}{- k - 4 - 1 + k}$
$\Rightarrow \frac{4 k - 2}{1 - 2 k} = \frac{6 - 4 k}{- 5}$
$\Rightarrow 120 k + 10 = 6 - 4 k - 12 k + 8 k^{2}$
$\Rightarrow 8 k^{2} + 4 k - 4 = 0$
$\Rightarrow 2 k^{2} + k - 1 = 0$
$\Rightarrow 2 k^{2} (k + 1) - 1 (k + 1) = 0$
$\Rightarrow (2 k - 1) (k + 1) = 0$
$\Rightarrow 2 k - 1 = 0 o r k + 1 = 0$
$\Rightarrow k = \frac{1}{2} o r k = - 1$

Suggest Corrections

Similar questions

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

Join BYJU'S Learning Program