The area of the triangle formed by the pair of straight lines $(a x+b y)^2-3(b x-a y)^2=0$ and $a x+b y+c=0$ is

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Aicharya · Accepted Answer

Given that, equation of pair of straight lines is $\begin{aligned} &(a x+b y)^2-3(b x-a y)^2=0 \\ & \Rightarrow a^2 x^2+b^2 y^2+2 a b x y \\ &-3\left(b^2 x^2+a^2 y^2-2 a b x y\right)=0 \\ & \Rightarrow \quad\left(a^2-3 b^2\right) x^2-\left(b^2-3 a^2\right) y^2+8 a b x y=0 \end{aligned}$ Let the equation of line be $y=m_1 x, y=m_2 x$. $\therefore \quad m_1+m_2=\frac{-4 a b}{b^2-3 a^2} \quad\left[\begin{array}{c}m_1+m_2=-\frac{2 h}{b} \\ \because m_1 m_2=\frac{a}{b}\end{array}\right]$ and $m_1 m_2=\frac{a^2-3 b^2}{b^2-3 a^2}$ Now, $\left(m_1-m_2\right)^2=\left(m_1+m_2\right)^2-4 m_1 m_2$ $\begin{aligned} & =\frac{16 a^2 b^2}{\left(b^2-3 a^2\right)^2}-\frac{4\left(a^2-3 b^2\right)}{\left(b^2-3 a^2\right)} \\ & =\frac{4}{\left(b^2-3 a^2\right)^2} \quad\left[4 a^2 b^2-\left(a^2-3 b^2\right)\left(b^2-3 a^2\right)\right] \\ & =\frac{4}{\left(b^2-3 a^2\right)^2}\left[4 a^2 b^2-\left(a^2 b^2-3 a^4\right.\right. \\ & \left.=\frac{4}{\left(b^2-3 a^2\right)^2}\left[-6 a^2 b^2+3 a^4+3 b^4 b^2\right)\right] \\ & =\frac{24}{\left(b^2-3 a^2\right)^2}\left(a^2-b^2\right)^2 \end{aligned}$

The intersecting point on the lines $y=m_1 x$, $y=m_2 x \text { and } a x+b y+c=0$ 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 $\Delta=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ \frac{-c}{a+b m_1} & \frac{-c m_1}{a+b m_1} & 1 \\ \frac{-c}{a+b m_2} & \frac{-c m_2}{a+b m_2} & 1\end{array}\right|$ $=\frac{1}{2}\left[\frac{+c^2 m_2}{\left(a+b m_1\right)\left(a+b m_2\right)}-\frac{c^2 m_1}{\left(a+b m_1\right)\left(a+b m_2\right)}\right]$ $=-\frac{1}{2}\left[\frac{c^2\left(m_1-m_2\right)}{a^2+a b\left(m_1+m_2\right)+b^2 m_1 m_2}\right]$ $\begin{aligned} & =\frac{-\frac{1}{2}\left[\frac{c^2 2 \sqrt{3}}{\left(b^2-3 a^2\right)} \cdot\left(a^2-b^2\right)\right]}{a^2+a b\left(\frac{-4 a b}{b^2-3 a^2}\right)+b^2\left(\frac{a^2-3 b^2}{b^2-3 a^2}\right)} \text { [from Eq. (i)] } \\ & =\frac{-\sqrt{3}\left[\left(a^2-b^2\right) c^2\right]}{a^2\left(b^2-3 a^2\right)-4 a^2 b^2+a^2 b^2-3 b^4} \\ & =\frac{-\sqrt{3}\left[c^2\left(a^2-b^2\right)\right]}{a^2 b^2-3 a^4-3 a^2 b^2-3 b^4} \\ & =\frac{-\sqrt{3} c^2}{-3\left(a^4+b^4+2 a^2 b^2\right)}=\frac{c^2}{\sqrt{3}\left(a^2+b^2\right)^2}\end{aligned}$

The area of the triangle formed by the pair of straight lines $(a x+b y)^2-3(b x-a y)^2=0$ and $a x+b y+c=0$…

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