Question

The area of the triangle formed by the pair of straight lines $(ax+by{)}^2-3(bx-ay{)}^2=0$ and $ax+by+c=0$ is

• A. $\frac{c^2}{a^2+b^2}$
• B. $\frac{c^2}{2\left(a^2+b^2\right)}$
• C. $\frac{c^2}{\sqrt{2}\left(a^2+b^2\right)}$
• D. $\frac{c^2}{\sqrt{3}\left(a^2+b^2\right)}$

Answer 1

Answer: (D)

Given that, equation of pair of straight lines is
$(ax+by{)}^2-3(bx-ay{)}^2=0$
$\Rightarrow a^2{x}^2+b^2{y}^2+2abxy$
$-3\left(b^2{x}^2+a^2{y}^2-2abxy\right)=0$
$\Rightarrow \left(a^2-3b^2\right)x^2-\left(b^2-3a^2\right)y^2+8abxy=0$
Let the equation of line be $y=m_{1}x,y=m_{2}x$.
$\therefore m_1+m_2=\frac{-4ab}{b^2-3a^2}$
$\left[\begin{array}{c}{m}_1+m_2=-\frac{2h}{b}\\ \because \\ m_1{m}_2=\frac{a}{b}\end{array}\right]$
and $m_1{m}_2=\frac{a^2-3b^2}{b^2-3a^2}$
Now, ${\left(m_1-m_2\right)}^2={\left(m_1+m_2\right)}^2-4m_1{m}_2$
$=\frac{16a^2{b}^2}{{\left(b^2-3a^2\right)}^2}-\frac{4\left(a^2-3b^2\right)}{\left(b^2-3a^2\right)}$
$=\frac{4}{{\left(b^2-3a^2\right)}^2}\left[4a^2{b}^2-\left(a^2-3b^2\right)\left(b^2-3a^2\right)\right]$
$=\frac{4}{{\left(b^2-3a^2\right)}^2}[4a^2{b}^2-(a^2{b}^2-3a^4-3b^4+9a^2{b}^2)]$
$=\frac{4}{{\left(b^2-3a^2\right)}^2}\left[-6a^2{b}^2+3a^4+3b^4\right]$
$=\frac{24}{{\left(b^2-3a^2\right)}^2}{\left(a^2-b^2\right)}^2$
$\therefore \left(m_1-m_2\right)=\frac{2\sqrt{3}}{\left(b^2-3a^2\right)}\cdot \left(a^2-b^2\right)...(i)$
The intersecting point on the lines $y=m_{1}x$,
$y=m_{2}x$ and $ax+by+c=0$
are $A(0,0),B\left(\frac{-c}{a+b{m}_1},\frac{-c{m}_2}{a+b{m}_1}\right)$
and $C\left(\frac{-c}{a+b{m}_2},\frac{-c{m}_1}{a+b{m}_2}\right)$
Area of $\Delta =\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ \frac{-c}{a+b{m}_1}& \frac{-c{m}_1}{a+b{m}_1}& 1\\ \frac{-c}{a+b{m}_2}& \frac{-c{m}_2}{a+b{m}_2}& 1\end{array}\right|$
$=\frac{1}{2}\left[\frac{+c^2{m}_2}{\left(a+b{m}_1\right)\left(a+b{m}_2\right)}-\frac{c^2{m}_1}{\left(a+b{m}_1\right)\left(a+b{m}_2\right)}\right]$
$=-\frac{1}{2}\left[\frac{c^2\left(m_1-m_2\right)}{a^2+ab\left(m_1+m_2\right)+b^2{m}_1{m}_2}\right]$
$=\frac{-\frac{1}{2}\left[\frac{c^{2}2\sqrt{3}}{\left(b^2-3a^2\right)}\cdot \left(a^2-b^2\right)\right]}{a^2+ab\left(\frac{-4ab}{b^2-3a^2}\right)+b^2\left(\frac{a^2-3b^2}{b^2-3a^2}\right)}[$ from Eq.(i)]
$=\frac{-\sqrt{3}\left[\left(a^2-b^2\right)c^2\right]}{a^2\left(b^2-3a^2\right)-4a^2{b}^2+a^2{b}^2-3b^4}$
$=\frac{-\sqrt{3}\left[c^2\left(a^2-b^2\right)\right]}{a^2{b}^2-3a^4-3a^2{b}^2-3b^4}$
$=\frac{-\sqrt{3}{c}^2}{-3\left(a^4+b^4+2a^2{b}^2\right)}$
$=\frac{c^2}{\sqrt{3}{\left(a^2+b^2\right)}^2}$