If ($\mathrm{\alpha }, \mathrm{\beta }$) is the centriod of the triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my = 1, prove that
$\frac{\mathrm{\alpha }}{\mathrm{bl}-\mathrm{hm}}=\frac{\mathrm{\beta }}{\mathrm{am}-\mathrm{hl}}=\frac{2}{3\left({\mathrm{bl}}^2-2\mathrm{hlm}+{\mathrm{am}}^2\right)}$

Let ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2=0$ represents the lines
$\begin{array}{l}{\mathrm{l}}_1\mathrm{x}+{\mathrm{m}}_1\mathrm{y}=0\dots ..\left(1\right){\mathrm{l}}_2\mathrm{x}+{\mathrm{m}}_2\mathrm{y}=0\dots .\left(2\right)\\ \therefore \left({\mathrm{l}}_1\mathrm{x}+{\mathrm{m}}_1\mathrm{y}\right)\left({\mathrm{l}}_2\mathrm{x}+{\mathrm{m}}_2\mathrm{y}\right)={\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2\end{array}$
Comparing like terms on  both sides we get
${\mathrm{l}}_1{\mathrm{l}}_2=\mathrm{a},{\mathrm{m}}_1{\mathrm{m}}_2=\mathrm{b},{\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{l}}_2{\mathrm{m}}_1=2\mathrm{h}$
Given line is lx + my = 1
Clearly the origin O is the point of intersection of (1) and (2)
Let A be the point of intersection of (1) and (3)
By the method of cross multiplication
$\begin{array}{clccc} \mathrm{x}& & y & 1 & \\ {\mathrm{m}}_1& 0& {\mathrm{l}}_1& {\mathrm{m}}_1& \\ \mathrm{m}& -1& l& \mathrm{m}& \end{array}$
$\begin{array}{l}\Rightarrow \frac{\mathrm{x}}{-{\mathrm{m}}_1-0}=\frac{\mathrm{y}}{0+{\mathrm{l}}_1}=\frac{1}{{\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1}\\ \Rightarrow \mathrm{x}=\frac{-{\mathrm{m}}_1}{{\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1},\mathrm{y}=\frac{{\mathrm{l}}_1}{{\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1}\\ \therefore \mathrm{A}=\left(\frac{-{\mathrm{m}}_1}{{\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1},\frac{{\mathrm{l}}_1}{{\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1}\right)\end{array}$
Similarly the point of intersection of (2) and (3) is
$\therefore \mathrm{B}=\left(\frac{-{\mathrm{m}}_2}{{\mathrm{l}}_2\mathrm{m}-{\mathrm{lm}}_2},\frac{{\mathrm{l}}_2}{{\mathrm{l}}_2\mathrm{m}-{\mathrm{lm}}_2}\right)$
Centroid of OAB is G = ($\mathrm{\alpha },\mathrm{\beta }$) =
$\begin{array}{l}\left(\frac{\frac{-{\mathrm{m}}_1}{{\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1}-\frac{-{\mathrm{m}}_2}{{\mathrm{l}}_2\mathrm{m}-{\mathrm{lm}}_2}}{3},\frac{\frac{{\mathrm{l}}_1}{{\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1}+\frac{{\mathrm{l}}_2}{{\mathrm{l}}_2\mathrm{m}-{\mathrm{lm}}_2}}{3}\right)\\ \therefore \mathrm{\alpha }=\frac{1}{3}\left[\frac{-{\mathrm{m}}_1}{{\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1}-\frac{{\mathrm{m}}_2}{{\mathrm{l}}_2\mathrm{m}-{\mathrm{lm}}_2}\right],\\ \mathrm{\beta }=\frac{1}{3}\left[\frac{{\mathrm{l}}_1}{{\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1}+\frac{{\mathrm{l}}_2}{{\mathrm{l}}_2\mathrm{m}-{\mathrm{lm}}_2}\right]\\ \therefore \mathrm{\alpha }=\frac{1}{3}\left[\frac{-{\mathrm{m}}_1\left({\mathrm{l}}_2\mathrm{m}-{\mathrm{lm}}_2\right)-{\mathrm{m}}_2\left({\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1\right)}{\left({\mathrm{l}}_1\mathrm{m}-{\mathrm{lm}}_1\right)\left({\mathrm{l}}_2\mathrm{m}-{\mathrm{lm}}_2\right)}\right]\\ =\frac{1}{3}\left[\frac{2{\mathrm{m}}_1{\mathrm{m}}_2\mathrm{l}-{\mathrm{l}}_2{\mathrm{m}}_1\mathrm{m}-{\mathrm{l}}_1{\mathrm{m}}_2\mathrm{m}}{{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}^2-\left({\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{l}}_2{\mathrm{m}}_1\right)\mathrm{lm}+{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{l}}^2}\right]\end{array}$
$\begin{array}{l}=\frac{2\mathrm{bl}-2\mathrm{hm}}{3\left({\mathrm{bl}}^2-2\mathrm{hlm}+{\mathrm{am}}^2\right)}\\ \Rightarrow \frac{\mathrm{\alpha }}{\mathrm{bl}-\mathrm{hm}}=\frac{2}{3\left({\mathrm{bl}}^2-2\mathrm{hlm}+{\mathrm{am}}^2\right)}\end{array}$
Similarly $\Rightarrow \frac{\mathrm{\beta }}{\mathrm{am}-\mathrm{hl}}=\frac{2}{3\left({\mathrm{bl}}^2-2\mathrm{hlm}+{\mathrm{am}}^2\right)}$
$\therefore \frac{\mathrm{\alpha }}{\mathrm{bl}-\mathrm{hm}}=\frac{\mathrm{\beta }}{\mathrm{am}-\mathrm{hl}}=\frac{2}{3\left({\mathrm{bl}}^2-2\mathrm{hlm}+{\mathrm{am}}^2\right)}$