Show  that the following lines form an equilateral triangle and find the area of the triangle
$\left(\mathrm{i}\right) {\mathrm{x}}^2-4\mathrm{xy}+{\mathrm{y}}^2=0,\mathrm{x}+\mathrm{y}=3$
$\text{ ii) }(\mathrm{x}+2\mathrm{a}{)}^2-3{\mathrm{y}}^2=0,\mathrm{x}=\mathrm{a}$

(i) Given line x + y – 3 = 0
Let any line through origin forming an equilateral triangle with the given line be
y = mx
To show the combined equation of such lines
$\overline{\mathrm{OA}}\text{ and }\overline{\mathrm{OB}}\text{ as }{\mathrm{x}}^2-4\mathrm{xy}+{\mathrm{y}}^2=0$
$\text{ Let }\mathrm{\angle }\mathrm{OAB}={60}^{\circ }$
Let M₁ = slope of line through (0, 0) = m
${\mathrm{M}}_2=\text{ slope of given line }=\frac{-\mathrm{a}}{\mathrm{b}}=\frac{-1}{1}=-1$
Since the angle between lines is 60°.

$$\begin{gathered} \text{ So }\mathrm{Tan}\mathrm{\theta }=\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right| \\ \Rightarrow \mathrm{Tan}{60}^0=\left|\frac{\mathrm{m}+1}{1-\mathrm{m}}\right| \end{gathered}$$
$\sqrt{3}=\left|\frac{{\mathrm{m}}_{ }+1}{1-\mathrm{m}}\right|$
s.o.b.s.
$\begin{array}{l}3={\left(\frac{\mathrm{m}+1}{1-\mathrm{m}}\right)}^2\\ \Rightarrow 3={\left(\frac{\mathrm{y}/\mathrm{x}+1}{1-\mathrm{y}/\mathrm{x}}\right)}^2\left[\because \mathrm{m}=\frac{\mathrm{y}}{\mathrm{x}}\right]\\ 3={\left(\frac{\mathrm{y}+\mathrm{x}}{\mathrm{x}-\mathrm{y}}\right)}^2\\ 3(\mathrm{x}-\mathrm{y}{)}^2=(\mathrm{y}+\mathrm{x}{)}^2\\ 3\left({\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}\right)={\mathrm{y}}^2+{\mathrm{x}}^2+2\mathrm{xy}\\ 3{\mathrm{x}}^2+3{\mathrm{y}}^2-6\mathrm{xy}={\mathrm{y}}^2+{\mathrm{x}}^2+2\mathrm{xy}\\ 2{\mathrm{x}}^2-8\mathrm{xy}+2{\mathrm{y}}^2=0\\ {\mathrm{x}}^2-4\mathrm{xy}+{\mathrm{y}}^2=0\end{array}$
Hence the given lines form an equilateral triangle.
$\text{ ii) Given }(\mathrm{x}+2\mathrm{a}{)}^2-3{\mathrm{y}}^2=0$
$\begin{array}{l}\Rightarrow (\mathrm{x}+2\mathrm{a}{)}^2-(\sqrt{3}\mathrm{y}{)}^2=0\\ \Rightarrow (\mathrm{x}+2\mathrm{a}+\sqrt{3}\mathrm{y})(\mathrm{x}+2\mathrm{a}-\sqrt{3}\mathrm{y})=0\end{array}$
The two lines are $\mathrm{x}+\sqrt{3}\mathrm{y}+2\mathrm{a}=0\dots \left(1\right)$
$\mathrm{x}-\sqrt{3}\mathrm{y}+2\mathrm{a}=0 ..\left(2\right)$
$\text{ Given line is }\mathrm{x}-\mathrm{a}=0\dots \left(3\right)$
‘A’ is point of intersection of (1) and (2)
A = (–2a, 0)
‘B’ is point of intersection of (2) and (3)
$\therefore$B(a, $\sqrt{3}$a)
‘C’ is point of intersection of (3) and (1)
C ( a, -$\sqrt{3}$a )
$\text{ Now }\mathrm{AB}=\sqrt{(\mathrm{a}+2\mathrm{a}{)}^2+(\sqrt{3}\mathrm{a}-0{)}^2}$
$\begin{array}{l}=\sqrt{9{\mathrm{a}}^2+3{\mathrm{a}}^2}=\sqrt{12{\mathrm{a}}^2}\\ \mathrm{BC}=\sqrt{(\mathrm{a}-\mathrm{a}{)}^2+(-\sqrt{3}\mathrm{a}-\sqrt{3}\mathrm{a}{)}^2}\\ =\sqrt{0+(-2\sqrt{3}\mathrm{a}{)}^2}=\sqrt{4\times 3{\mathrm{a}}^2}=\sqrt{12{\mathrm{a}}^2}\\ \mathrm{AC}=\sqrt{(\mathrm{a}+2\mathrm{a}{)}^2+(-\sqrt{3}\mathrm{a}-0{)}^2}\\ =\sqrt{9{\mathrm{a}}^2+3{\mathrm{a}}^2}=\sqrt{12{\mathrm{a}}^2}\\ \therefore \mathrm{AB}=\mathrm{BC}=\mathrm{AC}\end{array}$
The lines forms an equilateral triangle
$\text{ Area }=\frac{\sqrt{3}}{4}(\text{ side }{)}^2=\frac{\sqrt{3}}{4}\times 12{\mathrm{a}}^2$
$=3\sqrt{3}{\mathrm{a}}^2\text{ sq.units }$