If the points $(k, 2 - 2 k), (1 - k, 2 k)$ and $(- k - 4, 6 - 2 k)$ are collinear, possible values of $k$ are

$k = 2$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$k = - 1$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$k = 1$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$k = \frac{1}{2}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
B
$k = - 1$

D
$k = \frac{1}{2}$

Points $A (k, 2 - 2 k), B (1 - k, 2 k)$ and $C (- k - 4, 6 - 2 k)$ are collinear.
Then, slope of line AB = slope of line BC
$\frac{2 k - (2 - 2 k)}{1 - k - k} = \frac{6 - 2 k - 2 k}{- k - 4 - 1 + k}$
$\frac{4 k - 2}{1 - 2 k} = \frac{6 - 4 k}{- 5}$
$- 20 k + 10 = 6 - 4 k - 12 k + 8 k^{2}$
$8 k^{2} + 4 k - 4 = 0$
$2 k^{2} + k - 1 = 0$

$2 k^{2} + 2 k - k - 1$ = 0
$2 k (k + 1) - 1 (k + 1) = 0$
$(2 k - 1) (k + 1) = 0$
$2 k - 1 = 0 a n d k + 1 = 0$
$k = \frac{1}{2} a n d k = - 1$
we can also say, slope of AB = slope of AC $\frac{2 k - (2 - 2 k)}{1 - k - k} = \frac{6 - 2 k - (2 - 2 k)}{- k - 4 - k}$
$\frac{4 k - 2}{1 - 2 k} = \frac{4}{- 2 k - 4}$
$- 8 k^{2} + 4 k - 16 k + 8 = 4 - 8 k$
$8 k^{2} + 4 k - 4 = 0$
$2 k^{2} + k - 1 = 0$
We see that we obtained the same equation as in the previous case.
So, $k = \frac{1}{2}$ and $k = - 1$ are the only possible values of $k$ when given points are collinear.

Suggest Corrections