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Question: If three points \(\left( {k,2k} \right),\left( {2k,3k} \right),\left( {3,1} \right)\) are collinear,...

If three points (k,2k),(2k,3k),(3,1)\left( {k,2k} \right),\left( {2k,3k} \right),\left( {3,1} \right) are collinear, then k is equal to:
A. −2 B. 1 C. 12 D. −12  {\text{A}}{\text{. }} - 2 \\\ {\text{B}}{\text{. }}1 \\\ {\text{C}}{\text{. }}\dfrac{1}{2} \\\ {\text{D}}{\text{. }} - \dfrac{1}{2} \\\

Explanation

Solution

Hint- Here, we will be using the formula for slope between two points.

Let the given three points be P(k,2k),Q(2k,3k),R(3,1){\text{P}}\left( {k,2k} \right),{\text{Q}}\left( {2k,3k} \right),{\text{R}}\left( {3,1} \right)
As we know that the slope of any line joining two points A(a,b){\text{A}}\left( {a,b} \right) and B(c,d){\text{B}}\left( {c,d} \right) is given by m=d−bc−am = \dfrac{{d - b}}{{c - a}}.
Slope of line PQ is mPQ=3k−2k2k−k=kk=1{m_{PQ}} = \dfrac{{3k - 2k}}{{2k - k}} = \dfrac{k}{k} = 1
Slope of line QR is mQR=1−3k3−2k{m_{QR}} = \dfrac{{1 - 3k}}{{3 - 2k}}
Slope of line PR is mPR=1−2k3−k{m_{PR}} = \dfrac{{1 - 2k}}{{3 - k}}
For all the three given points to be collinear, the slope of lines made by joining any two points will be equal.
⇒mPQ=mQR=mPR⇒1=1−3k3−2k=1−2k3−k\Rightarrow {m_{PQ}} = {m_{QR}} = {m_{PR}} \Rightarrow 1 = \dfrac{{1 - 3k}}{{3 - 2k}} = \dfrac{{1 - 2k}}{{3 - k}}
When taking ⇒1=1−3k3−2k⇒3−2k=1−3k⇒k=−2 \Rightarrow 1 = \dfrac{{1 - 3k}}{{3 - 2k}} \Rightarrow 3 - 2k = 1 - 3k \Rightarrow k = - 2
And when ⇒1=1−2k3−k⇒3−k=1−2k⇒k=−2 \Rightarrow 1 = \dfrac{{1 - 2k}}{{3 - k}} \Rightarrow 3 - k = 1 - 2k \Rightarrow k = - 2
Therefore, the value of kk for which the given three points are collinear is −2 - 2.
Hence, option A is correct.

Note- In these types of problems where collinearity of points is the condition, the key feature is to make the slope of the lines joining any two points equal in order to find the unknown. In the above problem, while considered different slopes equality the value of the unknown comes out to be the same.