Question

If the points $\left( {k,2 - 2k} \right)$ $\left( {1 - k,2k} \right)$ and $\left( { - k - 4,6 - 2k} \right)$ be collinear the possible values of $k$ are
A. $- \dfrac{1}{2}$
B. $\dfrac{1}{2}$
C. $1$
D. $- 1$

Answer 1

Hint : We are asked to find the value of $k$ and an information is given that the points are collinear. To solve this question, first recall what collinear points are and what can be the conditions for these three points to be collinear. Use that condition and form equations to find the value of $k$ .

Complete step-by-step answer :
Given, the points $P\left( {k,2 - 2k} \right)$ , $Q\left( {1 - k,2k} \right)$ and $R\left( { - k - 4,6 - 2k} \right)$ .
It is also given that these points are collinear.
Let us first know the meaning of collinear points. The points which lie on the same line are known as collinear points.
The condition for collinear points: If $({x_1},{y_1}),\,({x_2},{y_2}),\,({x_3},{y_3})$ are three collinear points, then the slope of any two pairs of points will be the same.
Slope of a line segment joining the points $({x_1},{y_1})$ and $({x_2},{y_2})$ is,
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ (i)
We take two pairs of points that are two line segments, $PR$ and $QR$ and slopes of both line segments are equal.
For line segment $PR$ , we calculate the slope with the points $P\left( {k,2 - 2k} \right)$ and $R\left( { - k - 4,6 - 2k} \right)$ .
Here, ${y_2} = 6 - 2k$ and ${y_1} = 2 - 2k$
${x_2} = - k - 4$ and ${x_1} = k$
Putting these values in equation (i), we get
$m = \dfrac{{\left( {6 - 2k} \right) - \left( {2 - 2k} \right)}}{{\left( { - k - 4} \right) - k}}$
$\Rightarrow m = \dfrac{{6 - 2k - 2 + 2k}}{{ - k - 4 - k}}$
$\Rightarrow m = \dfrac{4}{{ - 2k - 4}}$
$\Rightarrow m = - \dfrac{2}{{k + 2}}$ (ii)
For line segment $QR$ , we calculate the slope with the points $Q\left( {1 - k,2k} \right)$ and $R\left( { - k - 4,6 - 2k} \right)$ .
Here, ${y_2} = 6 - 2k$ and ${y_1} = 2k$
${x_2} = - k - 4$ and ${x_1} = 1 - k$
Putting these values in equation (i), we get
$m = \dfrac{{\left( {6 - 2k} \right) - \left( {2k} \right)}}{{\left( { - k - 4} \right) - \left( {1 - k} \right)}}$
$\Rightarrow m = \dfrac{{6 - 2k - 2k}}{{ - k - 4 - 1 + k}}$
$\Rightarrow m = \dfrac{{6 - 4k}}{{ - 5}}$
$\Rightarrow m = - \dfrac{{6 - 4k}}{5}$ (iii)
Now, equating equations (ii) and (iii) we get,
$m = - \dfrac{2}{{k + 2}} = - \dfrac{{6 - 4k}}{5}$
$\Rightarrow \dfrac{2}{{k + 2}} = \dfrac{{6 - 4k}}{5}$
$\Rightarrow 10 = \left( {6 - 4k} \right)\left( {k + 2} \right)$
$\Rightarrow 10 = 6k + 12 - 4{k^2} - 8k$
$\Rightarrow 10 = - 4{k^2} - 2k + 12$
$\Rightarrow 4{k^2} + 2k - 2 = 0$
$\Rightarrow 2{k^2} + k - 1 = 0$
$\Rightarrow 2{k^2} + 2k - k - 1 = 0$
$\Rightarrow 2k\left( {k + 1} \right) - 1\left( {k + 1} \right) = 0$
$\Rightarrow \left( {2k - 1} \right)\left( {k + 1} \right) = 0$
$\Rightarrow \left( {2k - 1} \right)\, = 0\,{\text{or}}\left( {k + 1} \right) = 0$
$\Rightarrow k\, = \dfrac{1}{2}\,{\text{or}}\,k = - 1$
Therefore, for the points to be collinear there can be two values of $k$ these are $\dfrac{1}{2}$ or $- 1$
Hence, there can be correct answers , option (B) $\dfrac{1}{2}$ and option (D) $- 1$ .
So, the correct answer is “Option B and D”.

Note : On the basis of position, there are two types of points; these are collinear points and non-collinear points. Collinear points as we have discussed here, are the set of points which lie on the same line whereas non collinear points are the sets of points which do not lie on the same line. In case of non collinear points, the slope of any two pairs of points will not be the same. If we are given a set of points and are asked to determine whether they are collinear or non collinear points then we can check the slopes of any two pairs of the given set and know whether they are collinear or non collinear points.