Question

Find the value of \[k\] if the points \[A\left( {2,3} \right),B\left( {4,k} \right){\text{ and }}\left( {6, - 3} \right)\] are collinear.

Answer 1

Hint: The given three points are collinear if the area of the triangle formed by them is equal to zero. The area of the triangle formed by the points A \[\left( {{x_1},{y_1}} \right)\], B \[\left( {{x_2},{y_2}} \right)\] and C \[\left( {{x_3},{y_3}} \right)\] is given by \[\Delta = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:

Given points are \[A\left( {2,3} \right),B\left( {4,k} \right){\text{ and }}\left( {6, - 3} \right)\]
If the above points are collinear, they will lie on the same line i.e., they will not form a triangle.

Therefore, area of \[\Delta ABC = 0\]

We know that the area of the triangle formed by the points A \[\left( {{x_1},{y_1}} \right)\], B \[\left( {{x_2},{y_2}} \right)\] and C \[\left( {{x_3},{y_3}} \right)\] is given by \[\Delta = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]

So, area of \[\Delta ABC\] is given by

\[

  \dfrac{1}{2}\left[ {2\left( {k - \left( { - 3} \right)} \right) + 4\left( { - 3 - 3} \right) + 6\left( {3 - k} \right)} \right] = 0 \\

  2\left( {k + 3} \right) + 4\left( { - 6} \right) + 6\left( {3 - k} \right) = 0 \times 2 \\

  2k + 6 - 24 + 18 - 6k = 0 \\

  2k - 6k = - 6 + 24 - 18 \\

   - 4k = 0 \\

  \therefore k = 0 \\

\]

Thus, the value of \[k\] is 0.

Note: We can prove the collinearity of the three by another method i.e., if the sum of the lengths of any two-line segments among AB, BC, and CA is equal to the length of the remaining line segment, then the points are said to be in collinear. By these methods also we can find the value of \[k\].