Using distances alone, decide whether the points A(1, 2), B(4, 6) and C(10, 14) lie on one straight line.

$AB=\sqrt{(4-1{)}^2+(6-2{)}^2}=\sqrt{9+16}=5,BC=\sqrt{(10-4{)}^2+(14-6{)}^2}=\sqrt{36+64}=10$ and $AC=\sqrt{(10-1{)}^2+(14-2{)}^2}=\sqrt{81+144}=15$. Since $5+10=15$, the shortest route from A to $C$ passes through $B$, so the three points are collinear with $B$ between A and $C$. Equal distances are not required for collinearity.