Find the distance between the points:

Solution:

(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.