Q.Using the distance formula, show that the points $A(1, 2)$, $B(3, 8)$ and $C(4, 11)$ are collinear.

Three points are collinear if the distance between the two outer points equals the sum of the distances from each outer point to the one in between, so compute all three pairwise distances and check this addition.

$AB=\sqrt{(3-1)^2+(8-2)^2}=\sqrt{4+36}=\sqrt{40}=2\sqrt{10}$.

$BC=\sqrt{(4-3)^2+(11-8)^2}=\sqrt{1+9}=\sqrt{10}$. …

Three points are collinear exactly when the distance between the two outermost points equals the sum of the distances from each of them to the middle point. Compute all three distances using $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

$AB$ (from $A(1,2)$ to $B(3,8)$):

$BC$ (from $B(3,8)$ to $C(4,11)$):

$AC$ (from $A(1,2)$ to $C(4,11)$):

Now check whether $AB+BC=AC$:

$AB+BC = 2\sqrt{10}+\sqrt{10} = 3\sqrt{10} = AC$ …

An alternative method is to check that the slope of $AB$ equals the slope of $BC$ (both equal $3$ here, since $y$ increases by $3$ for every $1$ unit increase in $x$ along $A\to B\to C$); equal slopes through a common point also prove …

A frequent error is checking only whether $AB=BC+AC$ or $BC=AB+AC$ (i.e. testing the wrong point as the "middle" one) instead of correctly identifying, from the coordinates, w …