Find the distance between the points:
(i) $A(2,-3)$ and $B(-6,3)$
(ii) $C(-1,-1)$ and $D(8,11)$
(iii) $P(-8,-3)$ and $Q(-2,-5)$
(iv) $R(a+b, a-b)$ and $S(a-b, a+b)$
(i) Formula Used:
Distance between any two points $\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ and $\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=$
$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
Distance between $A(2,-3)$ and $B(-6,3)$
$=\sqrt{(-6-2)^{2}+(3-(-3))^{2}}$
$=\sqrt{64+36}=\sqrt{100}$
$=10$ units
Therefore, the distance between points A and B is 10 units.
(ii) Distance between $C(-1,-1)$ and $D(8,11)=$
$\sqrt{(8-(-1))^{2}+(11-(-1))^{2}}$
$=\sqrt{81+144}=\sqrt{225}$
$=15$ units
Therefore, the distance between points C and D is 10 units.
(iii) Distance between $P(-8,-3)$ and $Q(-2,-5)=$
$\sqrt{(-2-(-8))^{2}+(-5-(-3))^{2}}$
$=\sqrt{36+4}=\sqrt{40}$
$=2 \sqrt{10}$ units
Therefore, the distance between the points $P$ and $Q$ is $2 \sqrt{10}$ units.
(iv) Distance between R(a + b, a - b) and S(a - b, a +
b) $\sqrt{((a-b)-(a+b))^{2}+((a+b)-(a-b))^{2}}$
$=\sqrt{4 b^{2}+4 b^{2}}$
$=2 b \sqrt{2}$ units
Therefore, the distance between the points $\mathrm{R}$ and $\mathrm{S}$ is $2 \mathrm{~b} \sqrt{2}$ units.