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 value(s) of $k$ is/are
A.$- \dfrac{1}{2}$
B.$\dfrac{1}{2}$
C.$1$
D.$- 2$

Answer 1

Here we get to find the values of $k$. We will use the condition of collinearity to find the values of $k$. Collinear points are the points that lie on the same line. If two or more than two points lie on a line close to or far from each other, then they are said to be collinear.
Formula used:
Slope of the line segment joining two points say $({x_1},{\rm{ }}{y_1})$ and $({x_2},{\rm{ }}{y_2})$ is given by the formula: $m = \dfrac{{({y_2}-{y_1})}}{{({x_2} - {x_1})}}$ .

Complete step-by-step answer:
We are given that the points are collinear.
Now, taking the points $\left( {k,2 - 2k} \right)$and $\left( {1 - k,2k} \right)$ and by using the formula $m = \dfrac{{({y_2}-{y_1})}}{{({x_2} - {x_1})}}$, we get
${m_1} = \dfrac{{2k - 2 + 2k}}{{1 - k - k}} = \dfrac{{4k - 2}}{{1 - 2k}}$ ………………………………….(1)
Now, taking the points $\left( {1 - k,2k} \right)$and $\left( { - k - 4,6 - 2k} \right)$ and by using the formula $m = \dfrac{{({y_2}-{y_1})}}{{({x_2} - {x_1})}}$, we get
${m_2} = \dfrac{{6 - 2k - 2k}}{{ - k - 4 - 1 + k}} = \dfrac{{6 - 4k}}{{ - 5}}$ ……………………………….. (2)
Now, taking the points $\left( {k,2 - 2k} \right)$and $\left( { - k - 4,6 - 2k} \right)$ and by using the formula $m = \dfrac{{({y_2}-{y_1})}}{{({x_2} - {x_1})}}$, , we get
${m_3} = \dfrac{{6 - 2k - 2 + 2k}}{{ - k - 4 - k}} = \dfrac{4}{{ - 2k - 4}}$ …………………………….(3)
Since the points are collinear, the slope of all the points must be equal. So,
${m_1} = {m_2}$
Substituting the values of ${m_1}$ and ${m_2}$, we get
$\Rightarrow \dfrac{{4k - 2}}{{1 - 2k}} = \dfrac{{6 - 4k}}{{ - 5}}$
By cross-multiplication, we get
$\Rightarrow \left( {4k - 2} \right)\left( { - 5} \right) = \left( {6 - 4k} \right)\left( {1 - 2k} \right)$
Multiplying the terms, we get
$\Rightarrow - 20k + 10 = 6 - 12k - 4k + 8{k^2}$
Rewriting the equation, we get
$\Rightarrow 8{k^2} - 4k - 12k + 20k + 6 - 10 = 0$
Adding and Subtracting the like terms, we get
$\Rightarrow 8{k^2} + 4k - 4 = 0$
Dividing the equation by $2$on both the sides, we get
$\Rightarrow 2{k^2} + k - 1 = 0$
By factorization, we get
$\Rightarrow 2{k^2} + 2k - k - 1 = 0$
$\Rightarrow 2k\left( {k + 1} \right) - 1\left( {k + 1} \right) = 0$
$\Rightarrow \left( {2k - 1} \right)\left( {k + 1} \right) = 0$
Using zero product property, we get
$\begin{array}{l} \Rightarrow \left( {2k - 1} \right) = 0\\\ \Rightarrow k = \dfrac{1}{2}\end{array}$
Or
$\begin{array}{l} \Rightarrow \left( {k + 1} \right) = 0\\\ \Rightarrow k = - 1\end{array}$
Therefore, the values of $k$ are $- 1$ and $\dfrac{1}{2}$.

Note: Here, we have equated equation (1) and (2) to find the value of $k$. The same values of $k$ can be obtained by equating and solving equation (2) and (3). Three or more points are said to be collinear if the slope of any two pairs of points is the same. This can also be proved using the area of triangle formula. If the area of a triangle formed by three points is zero, then they are said to be collinear.