Find the distance between the following pairs of points :

Solution:

(a) The given points are : A (2, 3), B (4, 1).

Required distance $=\mathrm{AB}=\mathrm{BA}=\sqrt{\left(\mathbf{x}_{2}-\mathbf{x}_{1}\right)^{2}+\left(\mathbf{y}_{2}-\mathbf{y}_{1}\right)^{2}}$

$A B=\sqrt{(4-2)^{2}+(1-3)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}$

$=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$ units

(b) Here $x_{1}=-5, y_{1}=7$ and $x_{2}=-1, y_{2}=3$

$\therefore$ The required distance

$=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{\mathbf{I}-\mathbf{1}-\mathbf{( - 5 )}^{\mathbf{2}}+\mathbf{( 3}-\mathbf{7 )}^{2}}$

$=\sqrt{(-1+5)^{2}+(-4)^{2}}$

$=\sqrt{16+16}=\sqrt{32}=\sqrt{2 \times 16}$

$=4 \sqrt{2}$ units

(c) Here $x_{1}=a, y_{1}=b$ and $x_{2}=-a, y_{2}=-b$

$\therefore$ The required distance

$=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(-\mathbf{a}-\mathbf{a})^{2}+(-\mathbf{b}-\mathbf{b})^{2}}$

$=\sqrt{(-2 a)^{2}+(-2 b)^{2}}=\sqrt{4 a^{2}+4 b^{2}}$

$=\sqrt{4\left(a^{2}+b^{2}\right)}=2 \sqrt{\left(a^{2}+b^{2}\right)}$ units