The area of the triangle formed by the pair of lines $23 x^2-$ $48 x y+3 y^2=0$ with the line $2 x+3 y+5=0$ is

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Aicharya · Accepted Answer

Given the equation $23 x^2-48 x y+3 y^2=0$ After solving this equation we get two lines $\begin{aligned} & \mathrm{L}_1: y=\left(\frac{24+13 \sqrt{3}}{3}ight) x \ & \mathrm{~L}_2: y=\left(\frac{24-13 \sqrt{3}}{3}ight) x 	ext { and } \mathrm{L}_3: y=\frac{-2 x-5}{3}\end{aligned}$ After solving $\left(\mathrm{L}_1, \mathrm{~L}_2ight),\left(\mathrm{L}_2, \mathrm{~L}_3ight)$ and $\left(\mathrm{L}_1, \mathrm{~L}_3ight)$ We get $(0,0),\left(\frac{-5}{26-13 \sqrt{3}}, \frac{65 \sqrt{3}-120}{3(26-13 \sqrt{3})}ight)$ and $\left(\frac{-5}{26+13 \sqrt{3}}, \frac{-65 \sqrt{3}-120}{3(26+13 \sqrt{3})}ight)$ So, required area $=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \ \frac{-5}{26-13 \sqrt{3}} & \frac{65 \sqrt{3}-120}{3(26-13 \sqrt{3})} & 1 \ \frac{-5}{26+13 \sqrt{3}} & \frac{-65 \sqrt{3}-120}{3(26+13 \sqrt{3})} & 1 \end{array}ight|$ $\begin{aligned} & =\frac{1}{2}\left\{\frac{325 \sqrt{3}+600}{3(676-507)}+\frac{325 \sqrt{3}-600}{3(676-507)}ight\} \ & =\frac{1}{2}\left\{\frac{650 \sqrt{3}}{3 	imes 169}ight\}=\frac{25 \sqrt{3}}{39}=\frac{25}{13 \sqrt{3}} 	ext { sq. unit }\end{aligned}$

The area of the triangle formed by the pair of lines $23 x^2-$ $48 x y+3 y^2=0$ with the line $2 x+3 y+5=0$ is

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