Question

The straight line $(A^2-3B^2)x^2+8ABxy+(B^2-3A^2)y^2=0$ forms with the line $Ax+By+C=0$ an equilateral triangle of area

• A. $\frac{c^2}{\sqrt{2}\times (A^2+B^2)}$
• B. $\frac{c^2}{\sqrt{3}\times (A^2+B^2)}$
• C. $\frac{c^2}{(A^2+B^2)}$
• D. none of these

Answer 1

Answer: (B)

$\frac{c^2}{\sqrt{3}\times (A^2+B^2)}$

Let $y=mx$ be any line through the origin.

It makes an angle of $\pm {60}^0$ with $Ax+By+C=0.$

$\therefore \tan (\pm {60}^0)=\frac{m+\frac{A}{B}}{1-m\times \frac{A}{B}}$ $\Rightarrow \pm \sqrt{3}=\frac{mB+A}{B-mA}$

$\therefore 3{(B-mA)}^2={(mB+A)}^2$

The combined equation of the lines $AB$ and $AC$ is obtained by putting $m=\frac{y}{x}$.

$\therefore 3{(B-\frac{y}{x}A)}^2={(B\times \frac{y}{x}+A)}^2$ $\Rightarrow 3{(Bx-Ay)}^2={(By+Ax)}^2$

which on simplifications reduces to the given equation and hence proved.

Now, if $P$ is perpendicular from $A$ on $BC$, then

$P=\frac{C}{\sqrt{(A^2+B^2)}}$ ---------1)

Area of $△=\frac{1}{2}\times \frac{BC}{AD}=\frac{1}{2}\left(2BD\right)\times AD$ $=P\left(\tan {30}^0\right)P$

$=\frac{1}{\sqrt{3}}{P}^2=\frac{C^2}{\sqrt{3}(A^2+B^2)}$ ------------ [By (1)]