Question

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

Answer 1

Let the point be

$A=(-4,6),B=(1,3/2),C=(2,3)$

if we can show that $AB+BC=AC$

Then it's proved that these $3$ points can't from a triangle

$AB=\sqrt{(-4-1{)}^2+(-6-3/2{)}^2}=\sqrt{25+\frac{225}{4}}$

$=\sqrt{dfrac3254}$

$BC=\sqrt{(1-2{)}^2+(3/2-3{)}^2}=\sqrt{1+9/4}=\sqrt{13/4}$

$AC=\sqrt{(-4-2{)}^2+(-6-3{)}^2}=\sqrt{36+81}=\sqrt{117}$

$AB+BC=\sqrt{\frac{325}{4}}+\sqrt{13/4}$

$={\left[{(\sqrt{\frac{325}{4}}+\sqrt{13/4})}^2\right]}^{1/2}$

$={(\frac{325}{4}+\frac{13}{4}+2\sqrt{\frac{325}{4}}.\sqrt{13/4})}^{1/2}$

$={(\frac{338}{4}+2\sqrt{\frac{{13}^2.25}{4^2}})}^{1/2}$

$={(\frac{169}{2}+\frac{\mathrm{2.13.5}}{4})}^{1/2}$

$={\left(\frac{234}{2}\right)}^{1/2}=(117{)}^{1/2}$

$=\sqrt{117}$

$=AC$ [proved]