 QUESTION

# How can you say that points $\left( {1, - 5} \right),\left( {3, - 2} \right),\left( {7,4} \right)$ are collinear?

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.

Let A $\left( {{x_1},{y_1}} \right) = \left( {1, - 5} \right)$

B $\left( {{x_2},{y_2}} \right) = \left( {3, - 2} \right)$

C $\left( {{x_3},{y_3}} \right) = \left( {7,4} \right)$

To prove that three points are in collinear, the area of the triangle formed by the points A, B, and C is equal to zero.

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 $\Delta = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$

By substituting the above data, we get

$\Rightarrow \Delta = \dfrac{1}{2}\left[ {1\left( { - 2 - 4} \right) + 3\left( {4 - \left( { - 5} \right)} \right) + 7\left( { - 5 - \left( { - 2} \right)} \right)} \right] \\ \Rightarrow \Delta = \dfrac{1}{2}\left[ {1\left( { - 6} \right) + 3\left( 9 \right) + 7\left( { - 3} \right)} \right] \\ \Rightarrow \Delta = \dfrac{1}{2}\left[ { - 6 + 27 - 21} \right] \\ \Rightarrow \Delta = \dfrac{1}{2}\left[ { - 27 + 27} \right] \\ \therefore \Delta = 0 \\$

Hence, the area of the triangle formed by the points A, B, and C is equal to zero.
Thus, we can say that points $\left( {1, - 5} \right),\left( {3, - 2} \right),\left( {7,4} \right)$ are collinear.

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.