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