Question

The points $\left( k-1, k+2 \right),$ $\left( k, k+1 \right),$ $\left( k+1, k \right)$ are collinear for
A.Any value of $k$
B.$k=\dfrac{-1}{2}$ only
C.no value of $k$
D.integral values of $k$ only

Answer 1

Hint: Three points are collinear means that the three points are on the same straight line. Find the equation of the line using two of the three points, which can be found using two-point form$\dfrac{y_2-y_1}{x_2-x_1}$. And then, satisfy the equation with the coordinates of the third point to get the required condition.

Complete step by step answer:
Let the three given points be
$\begin{align}
  & A=(k-1,k+1) \\
 & B=(k,k+1) \\
 & C=(k+1,k) \\
\end{align}$
Now we know that three given points are collinear if they lie on the same straight line.
Hence, to find the condition for which the three given points are collinear, we can find the equation of the straight line for two of these points, and the third line will lie on the same line.
So, we can choose any two points to find the equation of the straight line.
Let’s choose points A and B.
The equation of the line joining two points, $\left( x_{1}, y_{1} \right)$ and $\left( x_{2}, y_{2} \right)$ can be given using the two-point form of the equation of a straight line, which is $\dfrac{y_2-y_1}{x_2-x_1}$
For the given points A and B,
$\begin{align}
  & {{x}_{1}}=k-1 \\
 & {{y}_{1}}=k+2 \\
 & {{x}_{2}}=k \\
 & {{y}_{2}}=k+1 \\
\end{align}$
Hence, using these values of $x_1, x_2, y_1, y_2$ in the above formula the equation of the line joining the points A and B in two-point form, we get
\[\dfrac{y-(k+2)}{x-(k-1)}=\dfrac{(k+1)-(k+2)}{k-(k-1)}\]
\[\Rightarrow \dfrac{y-(k+2)}{x-(k-1)}=\dfrac{-1}{1}\]
The above equation, after cross multiplication, can be written as
\[\Rightarrow y-(k+2)=-\left( x-(k-1) \right)\]
Upon rearranging with y on the LHS and other terms on the RHS, we get
\[\begin{align}
  & \Rightarrow y=\left( k-1 \right)-x+\left( k+2 \right)\ \\
 & \Rightarrow y=2k+1-x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots (1) \\
\end{align}\]
Thus the equation of the straight line joining the points A and B is $y=2k+1-x$
For the given condition of the three points being collinear, point $C \left( k+1, k \right)$ should also lie on the same straight line.
Thus, the coordinates of point C should satisfy the equation of the line AB.
Substituting the coordinates $C \left( k+1, k \right)$ in equation (1), we get
$k=2k+1-\left( k+1 \right)$
Solving the RHS of this equation, the RHS also comes out to be k.
Hence, the point $C$ will always lie on the line joining A and B for all values of k.
Therefore, the three points are collinear for any value of k.
Thus the correct answer is option (a).

Note:Another method to solve the question can be by using the fact that the area of the triangle formed by three collinear points is always 0.
We know that, area of a triangle having its vertices as$\left( x_1, y_1 \right)$,$\left( x_1, y_1 \right)$,$\left( x_1, y_1 \right)$ is given as $\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$
In the given question,
$y_3 = k$
Thus the formula for area becomes
\[\dfrac{1}{2}\left[ \left( k-1 \right)\left( k+1-k \right)+\left( k \right)\left( k-\left( k+2 \right) \right)+\left( k+1 \right)\left( \left( k+2 \right)-\left( k+1 \right) \right) \right]\]
$\Rightarrow \dfrac{1}{2}\left[ \left( k-1 \right)\left( 1 \right)+\left( k \right)\left( -2 \right)+\left( k+1 \right)\left( 1 \right) \right]$
$\begin{align}
  & \Rightarrow \dfrac{1}{2}\left[ k-1-2k+k+1 \right] \\
 & =\dfrac{1}{2}\left( 0 \right)=0 \\
\end{align}$
Hence the area of the triangle with the three given points is independent of the value of k and is always 0. Thus, the three points are collinear for any value of k.