Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0\text{ is }$$\frac{\left|\mathrm{c}\right|}{\sqrt{(\mathrm{a}-\mathrm{b}{)}^2+4{\mathrm{h}}^2}}$ 

Given equation
$\mathrm{S}={\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$
Let the lines represented by S = 0 be
${\mathrm{l}}_1\mathrm{x}+{\mathrm{m}}_1\mathrm{y}+{\mathrm{n}}_1=0 ...\left(1\right)$
$\mathrm{and} {\mathrm{l}}_2\mathrm{x}+{\mathrm{m}}_2\mathrm{y}+{\mathrm{n}}_2=0\dots \left(2\right)$
$\begin{array}{l}\therefore {\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}\\ =\left({\mathrm{l}}_1\mathrm{x}+{\mathrm{m}}_1\mathrm{y}+{\mathrm{n}}_1\right)\left({\mathrm{l}}_2\mathrm{x}+{\mathrm{m}}_2\mathrm{y}+{\mathrm{n}}_2\right)\end{array}$
Comparing the coefficients of like terms we get
$\begin{array}{l}{\mathrm{l}}_1{\mathrm{l}}_2=\mathrm{a};{\mathrm{m}}_1{\mathrm{m}}_2=\mathrm{b};{\mathrm{n}}_1{\mathrm{n}}_2=\mathrm{c}\\ {\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{m}}_1{\mathrm{l}}_2=2\mathrm{h},{\mathrm{l}}_1{\mathrm{n}}_2+{\mathrm{n}}_1{\mathrm{l}}_2=2\mathrm{g},{\mathrm{m}}_1{\mathrm{n}}_2+{\mathrm{m}}_2{\mathrm{n}}_1=2\mathrm{f}\end{array}$
Perpendicular distance from origin to the line $\text{ (1) is }\frac{\left|{\mathrm{n}}_1\right|}{\sqrt{{\mathrm{l}}_1^2+{\mathrm{m}}_1^2}}$
Perpendicular distance from origin to the line (2) is $\frac{\left|{\mathrm{n}}_2\right|}{\sqrt{{\mathrm{l}}_2^2+{\mathrm{m}}_2^2}}$
Product of the perpendicular distance from (0, 0) to S = 0 is
$\begin{array}{l}\frac{\left|{\mathrm{n}}_1\right|}{\sqrt{{\mathrm{l}}_1^2+{\mathrm{m}}_1^2}}\frac{\left|{\mathrm{n}}_2\right|}{\sqrt{{\mathrm{l}}_2^2+{\mathrm{m}}_2^2}}\\ =\frac{\left|{\mathrm{n}}_1{\mathrm{n}}_2\right|}{\sqrt{{\mathrm{l}}_1^2{\mathrm{l}}_2^2+{\mathrm{m}}_1^2{\mathrm{l}}_2^2+{\mathrm{l}}_1^2{\mathrm{m}}_2^2+{\mathrm{m}}_1^2{\mathrm{m}}_2^2}}\\ =\frac{\left|\mathrm{c}\right|}{\sqrt{{\left({\mathrm{l}}_1{\mathrm{l}}_2-{\mathrm{m}}_1{\mathrm{m}}_2\right)}^2+2{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}_1{\mathrm{m}}_2+{\left({\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{m}}_1{\mathrm{l}}_2\right)}^2-2{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}_1{\mathrm{m}}_2}}\\ =\frac{\left|\mathrm{c}\right|}{\sqrt{(\mathrm{a}-\mathrm{b}{)}^2+(2\mathrm{h}{)}^2}}=\frac{\left|\mathrm{c}\right|}{\sqrt{(\mathrm{a}-\mathrm{b}{)}^2+4{\mathrm{h}}^2}}\end{array}$
Hence proved