Without working out any square roots, decide which is longer: the segment AB with A(1, 1) and B(6, 5), or the segment CD with C(-2, 3) and D(3, 7).

$A{B}^2=(6-1{)}^2+(5-1{)}^2=25+16=41$ and $C{D}^2=(3-(-2){)}^2+(7-3{)}^2=25+16=41$. The squared lengths are equal, so the lengths are equal, each $\sqrt{41}\approx 6.40$. Comparing squares is enough for any such question, because squaring preserves order for positive lengths. In fact $CD$ is just $AB$ shifted 3 to the left and 2 up, and shifting never changes a length.