Answer
$d(A, B)+d(B, C)=\quad 4\sqrt{5}+2\sqrt{5}\quad =d(A, C)$,
so, the points are collinear.
Work Step by Step
$d(A, B) = \sqrt{(3-(-1))^{2}+(11-3)^{2}}= \sqrt{4^{2}+8^{2}}$
$= \sqrt{16+64}= \sqrt{80}= 4\sqrt{5}$.
$d(B, C) = \sqrt{(5-3)^{2}+(15-11)^{2}} = \sqrt{2^{2}+4^{2}}$
$= \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.
$d(A, C)=\sqrt{(5-(-1))^{2}+(15-3)^{2}}=\sqrt{6^{2}+12^{2}}$
$=\sqrt{36+144}=\sqrt{180}=6\sqrt{5}$.
We have
$d(A, B)+d(B, C)=\quad 4\sqrt{5}+2\sqrt{5}\quad =d(A, C)$,
so, the points are collinear.
