66
Using distance formula, prove that the points $(1, 5), (2, 3)$ and $(3, 1)$ are collinear.
Show SolutionHide Solution↓
Solution: Let $A(1, 5), B(2, 3)$ and $C(3, 1)$ be the points
$AB = \sqrt{1^2 + (-2)^2} = \sqrt{5}$
$BC = \sqrt{1^2 + (-2)^2} = \sqrt{5}$
$AC = \sqrt{2^2 + (-4)^2} = \sqrt{20}$ or $2\sqrt{5}$
$\therefore AB + BC = AC$, therefore points $A, B$ and $C$ are collinear.
$AB = \sqrt{1^2 + (-2)^2} = \sqrt{5}$
$BC = \sqrt{1^2 + (-2)^2} = \sqrt{5}$
$AC = \sqrt{2^2 + (-4)^2} = \sqrt{20}$ or $2\sqrt{5}$
$\therefore AB + BC = AC$, therefore points $A, B$ and $C$ are collinear.