Question

Using section formulas show that points (-3,-1), (1, 3) and (-1,1) are collinear.

Answer 1

Hint – In this question use the section formula as if a point $\left( {{x_3},{y_3}} \right)$ divides a line segment with endpoints $({x_1},{y_1}){\text{ and }}\left( {{x_2},{y_2}} \right)$ such that they all are collinear than the ratio will always be 1: 1. Use this concept to get the answer.

Complete Step-by-Step solution:

Let $A = (x_1, y_1) = (-3, -1)\\
B = (x_2, y_2) = (1, 3)\\
C = (x_3, y_3) = (-1, 1)$
We have to prove these three points are collinear.
Proof –
If these three points are collinear then they must lie on the same line.
Suppose we have a point $\left( {{x_3},{y_3}} \right)$dividing a line segment with endpoints as $({x_1},{y_1}){\text{ and }}\left( {{x_2},{y_2}} \right)$into the ratio $m:n$ then the section formula helps finding out the coordinate $\left( {{x_3},{y_3}} \right)$thus, ${x_3} = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},{y_3} = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}$
Let us suppose that point C divides the line AB internally in the ratio (m : 1) as shown in figure.
Then according to section formula the coordinates of C is given as
$ \Rightarrow {x_3} = \dfrac{{m{x_2} + 1.{x_1}}}{{m + 1}},{y_3} = \dfrac{{m{y_2} + 1.{y_1}}}{{m + 1}}$
Then substitute all the values in this formula and calculate the value of m,
$ \Rightarrow - 1 = \dfrac{{m\left( 1 \right) + 1\left( { - 3} \right)}}{{m + 1}},1 = \dfrac{{m\left( 3 \right) + 1\left( { - 1} \right)}}{{m + 1}}$
Now simplify the above equation we have,
$ \Rightarrow - 1 = \dfrac{{m - 3}}{{m + 1}},1 = \dfrac{{3m - 1}}{{m + 1}}$
$ \Rightarrow - m - 1 = m - 3,{\text{ }}m + 1 = 3m - 1$
$ \Rightarrow 2m = 2,{\text{ }}2m = 2$
Therefore from both of the equation the value of m is same which is
$ \Rightarrow m = \dfrac{2}{2} = 1$
So the point C divides the line in the ratio (1 : 1).
Thus the points A, B and C are collinear.
Hence Proved.

Note – Collinear means that the points are lying on the same line, there can be other conditions as well to prove the points to be collinear and not just by method by section formula. If three points $({x_1},{y_1}),\left( {{x_2},{y_2}} \right){\text{ and }}\left( {{x_3},{y_3}} \right)$ are collinear than$\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0$.