Question

Prove that the points (2, 3), (-4, -6) and `(1, 3/2)` do not form a triangle.

Answer 1

The distance d between two points `(x_1,y_1)` and `(x_2,y_2)` is given by the formula

`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`

In any triangle the sum of lengths of any two sides need to be greater than the third side.

Here the three points are A(2, 3), B(-4, -6) and C`(1, 3/2)` 

Let us now find out the lengths of all the three sides of the given triangle.

`AB = sqrt((2 + 4)^2 + (3 + 6)^2)`

`= sqrt((6)^2 + (9)^2)`

`= sqrt(36 + 81)`

`AB = sqrt(117)`

`BC = sqrt((-4 -1)^2 + (-6 - 3/2)^2)`

`= sqrt((-5)^2 + ((-15)/2)^2)`

`= sqrt(25 + 225/4)`

`BC = sqrt(81.24)`

`AC = sqrt((2 - 1)^2 + (3 - 3/2)^2)`

` = sqrt((1)^2 + (3/2)^2)`

`= sqrt(1 + 9/4)`

`AC = sqrt(3.25)`

Here we see that, BC + AC not greater than AB

This is in violation of the basic property of any triangle to exist. Therefore these points cannot give rise to a triangle.

Hence we have proved that the given three points do not form a triangle.