(i) Let $\mathrm{A}(1,-1), \mathrm{B}(5,2)$ and $\mathrm{C}(9,5)$ be the give points. Then

$\mathrm{AB}=\sqrt{(5-1)^{2}+(2+1)^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5 \text { units }$

$\mathrm{BC}=\sqrt{(9-5)^{2}+(5-2)^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5 \text { units }$

$\mathrm{AC}=\sqrt{(9-1)^{2}+(5+1)^{2}}=\sqrt{8^{2}+6^{2}}=\sqrt{100}=10 \text { units }$

$\therefore \mathrm{AB}+\mathrm{BC}=(5+5)$ units $=10$ units $=\mathrm{AC}$

Hence, the given points are collinear

(ii) Let $\mathrm{A}(6,9), \mathrm{B}(0,1)$ and $\mathrm{C}(-6,-7)$ be the give points. Then

$\mathrm{AB}=\sqrt{(0-6)^{2}+(1-9)^{2}}=\sqrt{(-6)^{2}+(-8)^{2}}=\sqrt{100}=10 \text { units }$

$\mathrm{BC}=\sqrt{(-6-0)^{2}+(-7-1)^{2}}=\sqrt{(-6)^{2}+(-8)^{2}}=\sqrt{100}=10 \text { units }$

$\mathrm{AC}=\sqrt{(-6-6)^{2}+(-7-9)^{2}}=\sqrt{(-12)^{2}+(16)^{2}}=\sqrt{400}=20 \text { units }$

$\therefore \mathrm{AB}+\mathrm{BC}=(10+10)$ units $=20$ units $=\mathrm{AC}$

=> the given points are collinear