Question

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

Answer 1

Let the triangle be ABC where
A(2,3)
B(-4,-6)
C($1$, $\frac{3}{2}$)

AB =

$AB=\sqrt{(}-4-2{)}^2+(-6-3{)}^2$
$AB=\sqrt{(}36+81)$
$AB=\sqrt{1}27\approx 11.27$


BC =
$BC=\sqrt{(}1+4{)}^2+(\frac{3}{2}+6{)}^2$
$BC=\sqrt{(}25+\frac{225}{4})$
$BC=\sqrt{\frac{325}{4}}\approx 9.01$

AC =
$AC=\sqrt{(}1-2{)}^2+(\frac{3}{2}-3{)}^2$
$AC=\sqrt{(}1+\frac{9}{4})$
$AC=\sqrt{\frac{13}{4}}\approx 1.8$

In a triangle sum of two sides is always greater than third side,

AC + BC = 1.8+9.01 = 10.81
AB =11.27

AB > AC +BC
Hence it is not a triangle