Question

Find the value of \[k\]for which the points\[A\left( {3k - 1,k - 2} \right)\] \[B\left( {k,k - 7} \right)\]& \[\left( {k - 1, - k - 2} \right)\]are collinear.

Answer 1

Hint: Property of points lying on the same line is known as collinearity, and the set of points which satisfy this property is known as collinear points. In the case of two points, they are trivially collinear since two points form a line. The Coordinate of a vertex is the 2-dimensional representation of a point given as\[M\left( {x,y} \right)\], where \[x\]represents the x-coordinates and \[y\]represents the y-coordinates.
In this question, it is already been given that the points \[A\left( {3k - 1,k - 2} \right)\] \[B\left( {k,k - 7} \right)\]& \[\left( {k - 1, - k - 2} \right)\]are collinear and we need to determine the value of k satisfying this condition.

Complete step by step answer:
\[A\left( {3k - 1,k - 2} \right)\], \[B\left( {k,k - 7} \right)\], \[C\left( {k - 1, - k - 2} \right)\]
To check whether the given points are collinear, find the slope between each pair of the points,
\[A\left( {3k - 1,k - 2} \right) \to A\left( {{x_1},{y_1}} \right)\]
\[B\left( {k,k - 7} \right) \to B\left( {{x_2},{y_2}} \right)\]
\[C\left( {k - 1, - k - 2} \right) \to C\left( {{x_3},{y_3}} \right)\]

We know the slope of the line is the ratio of change in the y-axis to the change in the x-axis represented as\[m = \dfrac{{\vartriangle y}}{{\vartriangle x}}\]hence we can write
The slope of line AB
\[{m_{AB}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{k - 7 - \left( {k - 2} \right)}}{{k - \left( {3k - 1} \right)}} = \dfrac{{k - 7 - k + 2}}{{k - 3k + 1}} = \dfrac{{ - 5}}{{ - 2k + 1}}\]
The slope of line BC
 \[{m_{BC}} = \dfrac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}} = \dfrac{{ - k - 2 - \left( {k - 7} \right)}}{{k - 1 - k}} = \dfrac{{ - k - 2 - k + 7}}{{ - 1}} = \dfrac{{ - 2k + 5}}{{ - 1}}\]
For the line to be collinear
The slope of line AB = Slope of line BC
\[
  {m_{AB}} = {m_{BC}} \\
  \dfrac{{ - 5}}{{ - 2k + 1}} = \dfrac{{ - 2k + 5}}{{ - 1}} \\
 \]
Now solve for \[k\] by cross multiplying we get
\[
  \dfrac{{ - 5}}{{ - 2k + 1}} = \dfrac{{ - 2k + 5}}{{ - 1}} \\
  \left( { - 5} \right)\left( { - 1} \right) = \left( { - 2k + 5} \right)\left( { - 2k + 1} \right) \\
  5 = 4{k^2} - 2k - 10k + 5 \\
  4{k^2} - 12k = 0 \\
  k\left( {4k - 12} \right) = 0 \\
  4k - 12 = 0 \\
  k = \dfrac{{12}}{4} \\
   = 3 \\
 \]
Hence the value of \[k = 3\]
Note:
 To check the collinearity of points find the slope of each pair of points whose collinearity is to be checked. We can check if the given sets of points are collinear by finding the slope between the pairs of the points, which must be equal for each pair. The slope of the line is generally defined as the direction of a line; it is the ratio of change in the horizontal axis to the change in the horizontal axis.