Question

Show that the straight lines
$(A^2-3B^2)x^2+8ABxy+(B^2-3A^2)y^2=0$
form with the line $Ax+By+C=0$ an equilateral triangle of area $\frac{C^2}{\surd 3.(A^2+B^2)}$.

Answer 1

Let $m$ be the slops of the line $AB$ or $AC$ which
passes through $A(0,0)$ and hence their equations will be
of the type $y=mx$ and for both of them
$m=y/m$.
If the triangle is to be equilateral, then
$\angle B=\angle C={60}^o$. Slope of line $Ax+By+C=0$ is $-A/B$.
Now $AB$ and $AC$ make angles of ${60}^o$ and $-{60}^o$ respectively with $BC$.

$\therefore \tan (\pm {60}^o)=\frac{m+A/B}{1-m.A/B}$ or $\pm \surd 3=\frac{mB+A}{B-mA}$
$\therefore 3(B-mA{)}^2=(mB+A{)}^2$.
The combined equation of the lines $AB$ and $AC$ is
obtained by putting
$m=y/x$, from$\left(1\right)$

$\therefore 3{(B-\frac{y}{x}A)}^2={(B.\frac{y}{x}+A)}^2$
or $3(Bx-Ay{)}^2=(By+Ax{)}^2$ which on simplification
reduces to the given equation and hence proved.
Now if $p$ be perpendicular from $A$ on $BC$, then
$p=\frac{C}{\surd (A^2+B^2)}$
Area of $\Delta =\frac{1}{2}BC.AD$
$=\frac{1}{2}.\left(2BD\right).AD$
$=\left(p\tan {30}^o\right)p=\frac{1}{\surd 3}.p^2$
$=\frac{C^2}{\surd 3.(A^2+B^2)}$ by $\left(1\right)$