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

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

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

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 $C\left(1, \frac{3}{2}\right)$

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

$A B=\sqrt{(2+4)^{2}+(3+6)^{2}}$

$=\sqrt{(6)^{2}+(9)^{2}}$

 

$=\sqrt{36+81}$

$A B=\sqrt{117}$

$B C=\sqrt{(-4-1)^{2}+\left(-6-\frac{3}{2}\right)^{2}}$

$=\sqrt{(-5)^{2}+\left(\frac{-15}{2}\right)^{2}}$

 

$=\sqrt{25+\frac{225}{4}}$

$B C=\sqrt{81.25}$

$A C=\sqrt{(2-1)^{2}+\left(3-\frac{3}{2}\right)^{2}}$

$=\sqrt{(1)^{2}+\left(\frac{3}{2}\right)^{2}}$

$=\sqrt{1+\frac{9}{4}}$

$A C=\sqrt{3.25}$

Here we see that, $B C+A C \ngtr A B$

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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