If an equation \(\mathrm{hxy}+\mathrm{g} x+\mathrm{f} y+\mathrm{c}=0\) represents a pair of lines, then

Given equation of pair of lines is
\(\begin{aligned}
& h x y+g x+f y+c=0 \\
\therefore \quad & A=0, B=0, C=c, F=\frac{f}{2}, G=\frac{g}{2}, H=\frac{h}{2}
\end{aligned}\)
\(\begin{aligned} & \therefore \quad\left|\begin{array}{lll}A & H & G \\ H & B & F \\ G & F & C\end{array}\right|=0 \\ & \Rightarrow\left|\begin{array}{ccc}0 & \frac{\mathrm{~h}}{2} & \frac{\mathrm{~g}}{2} \\ \frac{\mathrm{~h}}{2} & 0 & \frac{\mathrm{f}}{2} \\ \frac{\mathrm{~g}}{2} & \frac{\mathrm{f}}{2} & \mathrm{c}\end{array}\right|=0\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{-h}{2}\left[\frac{\mathrm{ch}}{2}-\frac{\mathrm{gf}}{4}\right]+\frac{\mathrm{g}}{2}\left[\frac{\mathrm{hf}}{4}\right]=0 \\ & \Rightarrow \frac{-\mathrm{h}^2 \mathrm{c}}{4}+\frac{\mathrm{hgf}}{8}+\frac{\mathrm{ghf}}{8}=0 \\ & \Rightarrow \frac{-\mathrm{h}^2 \mathrm{c}}{4}+\frac{\mathrm{hgf}}{4}=0 \\ & \Rightarrow \mathrm{~h}^2 \mathrm{c}=\mathrm{hgf} \\ & \Rightarrow \mathrm{hc}=\mathrm{gf}\end{aligned}\)