If the second degree equation S = ax² + 2hxy + by² + 2gx + 2fy + c = 0 in two variables x and y represents a pair of straight lines, then show that
$$\begin{gathered} \text{ (i) }\mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0\text{ and } \\ \text{ (ii) }{\mathrm{h}}^2\ge \mathrm{ab},{\mathrm{g}}^2\ge \mathrm{ac}\text{ and }{\mathrm{f}}^2\ge \mathrm{bc} \end{gathered}$$

Let the equation S = 0 represents  two lines ${\mathrm{l}}_1\mathrm{x}+{\mathrm{m}}_1\mathrm{y}+{\mathrm{n}}_1=0\text{ and }{\mathrm{l}}_2\mathrm{x}+{\mathrm{m}}_2\mathrm{y}+{\mathrm{n}}_2=0$
Then
$\begin{array}{l}{\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}\equiv \left({\mathrm{l}}_1\mathrm{x}+{\mathrm{m}}_1\mathrm{y}+{\mathrm{n}}_1\right)\left({\mathrm{l}}_2\mathrm{x}+{\mathrm{m}}_2\mathrm{y}+{\mathrm{n}}_2\right)\\ { }_{ }\end{array}$
Equating the co-efficients of like terms on both sides
we get ${\mathrm{l}}_1{\mathrm{l}}_2=\mathrm{a},{\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{l}}_2{\mathrm{m}}_1=2\mathrm{h},{\mathrm{m}}_1{\mathrm{m}}_2=\mathrm{b}$
${\mathrm{l}}_1{\mathrm{n}}_2+{\mathrm{l}}_2{\mathrm{n}}_1=2\mathrm{g},{\mathrm{m}}_1{\mathrm{n}}_2+{\mathrm{m}}_2{\mathrm{n}}_1=2\mathrm{f},{\mathrm{n}}_1{\mathrm{n}}_2=\mathrm{c}$
i) Consider the product (2h) (2g) (2f)
$\begin{array}{l}=\left({\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{l}}_2{\mathrm{m}}_1\right)\left({\mathrm{l}}_1{\mathrm{n}}_2+{\mathrm{l}}_2{\mathrm{n}}_1\right)\left({\mathrm{m}}_1{\mathrm{n}}_2+{\mathrm{m}}_2{\mathrm{n}}_1\right)\\ ={\mathrm{l}}_1{\mathrm{l}}_2\left({{\mathrm{m}}_1}^2{\mathrm{n}}_2^2+{\mathrm{m}}_2^2{\mathrm{n}}_1^2\right)+{\mathrm{m}}_1{\mathrm{m}}_2\left({\mathrm{l}}_1^2{\mathrm{n}}_2^2+{\mathrm{l}}_2^2{\mathrm{n}}_1^2\right)\\ +{\mathrm{n}}_1{\mathrm{n}}_2\left({\mathrm{l}}_1^2{{\mathrm{m}}_2}^2+{\mathrm{l}}_2{\mathrm{m}}_1^2\right)+2{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{n}}_1{\mathrm{n}}_2\\ ={\mathrm{l}}_1{\mathrm{l}}_2\left[{\left({\mathrm{m}}_1{\mathrm{n}}_2+{\mathrm{m}}_2{\mathrm{n}}_1\right)}^2-2{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{n}}_1{\mathrm{n}}_2\right]\\ +{\mathrm{m}}_1{\mathrm{m}}_2\left[{\left({\mathrm{l}}_1{\mathrm{n}}_2+{\mathrm{l}}_2{\mathrm{n}}_1\right)}^2-2{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{n}}_1{\mathrm{n}}_2\right]\\ +{\mathrm{n}}_1{\mathrm{n}}_2\left[{\left({\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{l}}_2{\mathrm{m}}_1\right)}^2-2{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}_1{\mathrm{m}}_2\right]+2{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{n}}_1{\mathrm{n}}_2\\ =\mathrm{a}\left(4{\mathrm{f}}^2-2\mathrm{bc}\right)+\mathrm{b}\left[4{\mathrm{g}}^2-2\mathrm{ac}\right]+\\ \mathrm{c}\left(4{\mathrm{h}}^2-2\mathrm{ab}\right)+2\mathrm{abc}\\ \Rightarrow 8\mathrm{fgh}=4\left[{\mathrm{af}}^2+{\mathrm{bg}}^2+{\mathrm{ch}}^2-\mathrm{abc}\right]\\ \therefore \mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0\end{array}$
$\begin{array}{l} \mathrm{ii}) {\mathrm{h}}^2-\mathrm{ab}={\left(\frac{{\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{l}}_2{\mathrm{m}}_1}{2}\right)}^2-{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}_1{\mathrm{m}}_2\\ =\frac{{\left({\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{l}}_2{\mathrm{m}}_1\right)}^2-4{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}_1{\mathrm{m}}_2}{4}\\ {\mathrm{h}}^2-\mathrm{ab}=\frac{{\left({\mathrm{l}}_1{\mathrm{m}}_2-{\mathrm{l}}_2{\mathrm{m}}_1\right)}^2}{4}\ge 0\\ \therefore {\mathrm{h}}^2\ge \mathrm{ab}\end{array}$
Similarly we can prove ${\text{g}}^2\ge \mathrm{ac} \mathrm{and} {\mathrm{f}}^2\ge \mathrm{bc}$