The points (1,1),(−2,7)  and (3,−3)

Let A(1,1),B(−2,7)  and C(3,−3)  be the given points, then AB=(−2−1)2+(7−1)2=9+36=35 BC=(3+2)2+(−3−7)2=25+100=55 AC=(3−1)2+(−3−1)2=4+16=25 Clearly BC=AB+AC, ⇒  A,B,C are collinear.