The equation of the pair of lines passing thr | vedclass

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Question

(A) Let the equations of the lines passing through the origin be $y = m_1x$ and $y = m_2x$. These lines form an equilateral triangle with the line $3x

Vedclass · Accepted Answer

$39x^2 + 11y^2 - 96xy = 0$

Vedclass · Answer

(A) Let the equations of the lines passing through the origin be $y = m_1x$ and $y = m_2x$. These lines form an equilateral triangle with the line $3x + 4y - 5 = 0$,which has a slope of $m = -\frac{3}{4}$.Since the triangle is equilateral,the angle between each line and the given line is $60^{\circ}$.Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight|$,we have $	an 60^{\circ} = \left| \frac{m - m_i}{1 + m \cdot m_i} ight|$.$\sqrt{3} = \left| \frac{-\frac{3}{4} - m_i}{1 + (-\frac{3}{4})m_i} ight| = \left| \frac{-3 - 4m_i}{4 - 3m_i} ight|$.Squaring both sides: $3 = \frac{(3 + 4m_i)^2}{(4 - 3m_i)^2} \Rightarrow 3(16 - 24m_i + 9m_i^2) = 9 + 24m_i + 16m_i^2$.$48 - 72m_i + 27m_i^2 = 9 + 24m_i + 16m_i^2$.$11m_i^2 - 96m_i + 39 = 0$.Since $m_1$ and $m_2$ are the roots of this quadratic equation,the combined equation of the pair of lines $y^2 - (m_1+m_2)xy + m_1m_2x^2 = 0$ is obtained by substituting $m = \frac{y}{x}$:$11(\frac{y}{x})^2 - 96(\frac{y}{x}) + 39 = 0$.Multiplying by $x^2$,we get $11y^2 - 96xy + 39x^2 = 0$.

The equation of the pair of lines passing through the origin and forming an equilateral triangle with the line $3x + 4y - 5 = 0$ is

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