Question

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

Answer 1

$(A^2-{3B}^2)x^2+8ABxy+(B^2-3A^2)y^2=0$

Given equation is homogenous equation of second degree. So the point of intersection of lines is $(0,0)$

Angle Between the pair of straight lines $\tan \theta =\left|\frac{2\sqrt{h^2-AB}}{A+B}\right|$

$\tan \theta =\left|\frac{2\sqrt{16A^2{B}^2-(A^2-{3B}^2)(B^2-3A^2)}}{A^2-{3B}^2+B^2-3A^2}\right|\tan \theta =\left|\frac{2\sqrt{16A^2{B}^2-(A^2{B}^2-3A^4-3B^4+9A^2{B}^2)}}{-2(A^2+B^2)}\right|\tan \theta =\left|\frac{\sqrt{3(A^4+B^4+2A^2{B}^2)}}{A^2+B^2}\right|=\sqrt{3}\Rightarrow \theta ={60}^{\circ }$

Now,

$(A^2-{3B}^2)x^2+8ABxy+(B^2-3A^2)y^2=0A^2{x}^2-{3B}^2{x}^2+8ABxy+B^2{y}^2-3A^2{y}^2=0A^2{x}^2+B^2{y}^2+2ABxy+6ABxy-{3B}^2{x}^2-3A^2{y}^2=0{(Ax+By)}^2-3{(Bx-Ay)}^2=0{(Ax+By)}^2-{\left(\sqrt{3}\right(Bx-Ay\left)\right)}^2=0$

Factorising the equation

$(Ax+By+\sqrt{3}Bx-\sqrt{3}Ay)\left(\right(Ax+By-(\sqrt{3}Bx-\sqrt{3}Ay))=0(A+\sqrt{3}B)x+(B-\sqrt{3}A)y=0,(A-\sqrt{3}B)x+(B+\sqrt{3}A)y=0(A+\sqrt{3}B)x+(B-\sqrt{3}A)y=\mathrm{0.........}(i\left)\right(A-\sqrt{3}B)x+(B+\sqrt{3}A)y=\mathrm{0........}(ii)$

Given line is $Ax+By+c=\mathrm{0.........}\left(iii\right)$

Angle Between $\left(i\right)$ And $\left(iii\right)$ that is $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan \theta =\left|\frac{\frac{A+\sqrt{3}B}{\sqrt{3}A-B}-\frac{-A}{B}}{1+\left(\frac{A+\sqrt{3}B}{\sqrt{3}A-B}\right)\left(\frac{-A}{B}\right)}\right|=\left|\frac{\frac{AB+\sqrt{3}{A}^2+\sqrt{3}{A}^2-AB}{(\sqrt{3}A-B)B}}{\frac{\sqrt{3}AB-A^2-B^2-\sqrt{3}AB}{(\sqrt{3}A-B)B}}\right|=\left|\frac{\sqrt{3}(A^2+B^2)}{{-A}^2-B^2}\right|\tan \theta =\sqrt{3}\Rightarrow \theta ={60}^{\circ }$

Angle Between $\left(ii\right)$ And $\left(iii\right)$

using Angle sum property of triangle $=180-60-60=60$

All the Angles of triangle are equal .Hence proved that triangle is equilateral.

Area of equilateral triangle $=\frac{\sqrt{3}}{4}{A}^2=\frac{p^2}{\sqrt{3}}$

where $a$ is the length of side and $p$ is the length of prependicular

Length of prependicular from $(0,0)$ to $Ax+By+c=0$ $=\frac{A.0+B.0+c}{\sqrt{A^2+B^2}}=\frac{c}{\sqrt{A^2+B^2}}$

Hence Area $=\frac{1}{\sqrt{3}}\times {\left(\frac{c}{\sqrt{A^2+B^2}}\right)}^2=\frac{c^2}{\sqrt{3}(A^2+B^2)}$