Find the distance between the following pairs of points :

(i) (-3, 6) and (2, -6)

(ii) (-a, -b) and (a, b)

(iii) $\Big(\dfrac{3}{5},2\Big)$ and $\Big(-\dfrac{1}{5}, 1\dfrac{2}{5}\Big)$

(iv) $\Big({\sqrt3 +1},1\Big)$ and (0, $\sqrt{3}$)

(i) Let (-3, 6) = (x₁, y₁) and (2, -6) = (x₂, y₂)

⇒ Distance between the given points =

$\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}\\[1em] = \sqrt{(2 - (-3))^2 + (-6 - 6)^2}\\[1em] = \sqrt{(5)^2 + (-12)^2}\\[1em] = \sqrt{25 + 144}\\[1em] = \sqrt{169}\\[1em] = 13

Hence, distance between the given points is 13.

(ii) Let (-a, -b) = (x₁, y₁) and (a, b) = (x₂, y₂)

⇒ Distance between the given points =

$\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}\\[1em] = \sqrt{(a - (-a))^2 + (b - (-b))^2}\\[1em] = \sqrt{(2a)^2 + (2b)^2}\\[1em] = \sqrt{4a^2 + 4b^2}\\[1em] = 2\sqrt{a^2 + b^2}\\[1em]

Hence, distance between the given points is $2\sqrt{a^2 + b^2}$.

(iii) Let $\Big(\dfrac{3}{5},2\Big)$ = (x₁, y₁) and $\Big(-\dfrac{1}{5}, 1\dfrac{2}{5}\Big)$ = (x₂, y₂)

⇒ Distance between the given points =

$\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}\\[1em] = \sqrt{\Big(\dfrac{-1}{5} - \dfrac{3}{5} \Big)^2 + \Big(\Big(1\dfrac{2}{5} - 2 \Big)\Big)^2}\\[1em] = \sqrt{\Big(\dfrac{-4}{5}\Big)^2 + \Big(\dfrac{7-10}{5}\Big)^2}\\[1em] = \sqrt{\dfrac{16}{25} + \Big(\dfrac{-3}{5}\Big)^2}\\[1em] = \sqrt{\dfrac{16}{25} + \dfrac{9}{25}}\\[1em] = \sqrt{\dfrac{25}{25}}\\[1em] = \sqrt{1}\\[1em] = 1

Hence, distance between the given points is 1.

(iv) Let $\Big({\sqrt3 +1},1\Big)$ = (x₁, y₁) and (0, $\sqrt{3}$) = (x₂, y₂)

⇒ Distance between the given points =

$\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}\\[1em] = \sqrt{(0 - ({\sqrt3 +1}))^2 + (\sqrt3 - 1)^2}\\[1em] = \sqrt{(-{(\sqrt3 +1)})^2 + (\sqrt3 - 1)^2}\\[1em] = \sqrt{3 + 1 + 2\sqrt3 + 1 + 3 - 2\sqrt3}\\[1em] = \sqrt{8}\\[1em] = 2\sqrt2

Hence, distance between the given points is $2\sqrt2$ = 2.83.