The lines represented b the equations $23 x^2-48 x y+3 y^2=0$ and $2 x+3 y+4=0$ form

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Aicharya · Accepted Answer

(c) We have, $ \begin{aligned} & 23 x^2-48 x y+3 y^2=0 \ & \Rightarrow \quad 3 y^2-48 x y+23 x^2=0 \ & 	ext { Here, } \quad M_1+M_2=16 \ & 	ext { and } M_1 M_2=\frac{23}{3} \ & 	herefore \quad m_1-m_2=\sqrt{\left(m_1+m_2ight)^2-4\left(m_1 m_2ight)} \ & =\sqrt{(16)^2-4 	imes \frac{23}{3}}=\sqrt{256-\frac{92}{3}}=\sqrt{\frac{768-92}{3}}=\frac{26}{\sqrt{3}} \ & 	herefore \quad m_1=8+\frac{13}{\sqrt{3}} 	ext { and } m_2=8-\frac{13}{\sqrt{3}} \ & 	an 	heta=\left|\frac{m_1-m_2}{1+m_1 m_2}ight|=\left|\frac{\frac{26}{\sqrt{3}}}{1+\frac{23}{3}}ight|=\sqrt{3} \ & \end{aligned} $ We know that, than $60^{\circ}$ So, $ 	heta=60^{\circ} $ Slope of line $2 x+3 y+4=0$ is $-\frac{2}{3}$ Angle between line of slope $8+\frac{13}{\sqrt{3}}$ and $-\frac{2}{3}$ is $ \begin{aligned} 	an \alpha & =\frac{8+\frac{13}{\sqrt{3}}+\frac{2}{3}}{1+\left(8+\frac{13}{\sqrt{3}}ight)\left(\frac{2}{3}ight)}=\sqrt{3} \ 	herefore \quad 	heta & =60^{\circ} \ \Rightarrow \quad \alpha & =60^{\circ} \end{aligned} $ Hence, line from a equilateral triangle

The lines represented b the equations $23 x^2-48 x y+3 y^2=0$ and $2 x+3 y+4=0$ form

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