The lines represented by the equations 23x^2 | vedclass

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(B) The pair of lines represented by $23x^2 - 48xy + 3y^2 = 0$ are $y = m_1x$ and $y = m_2x$. Comparing with $ax^2 + 2hxy + by^2 = 0$,we have $a = 23,

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an equilateral triangle

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(B) The pair of lines represented by $23x^2 - 48xy + 3y^2 = 0$ are $y = m_1x$ and $y = m_2x$. Comparing with $ax^2 + 2hxy + by^2 = 0$,we have $a = 23, 2h = -48, b = 3$. Sum of slopes $m_1 + m_2 = -\frac{2h}{b} = \frac{48}{3} = 16$. Product of slopes $m_1m_2 = \frac{a}{b} = \frac{23}{3}$. Difference of slopes $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1m_2} = \sqrt{16^2 - 4(\frac{23}{3})} = \sqrt{256 - \frac{92}{3}} = \sqrt{\frac{676}{3}} = \frac{26}{\sqrt{3}}$. The angle $	heta$ between these lines is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1m_2}| = |\frac{26/\sqrt{3}}{1 + 23/3}| = |\frac{26/\sqrt{3}}{26/3}| = \sqrt{3}$. Thus,$	heta = 60^{\circ}$. The slope of the third line $2x + 3y + 4 = 0$ is $m_3 = -2/3$. Calculating the angles between the third line and the pair of lines,we find that the triangle formed is an equilateral triangle.

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

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