The lines represented by the equation ax^2 + | vedclass

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(C) The given equation represents a pair of straight lines. Let the lines be $L_1: l_1x + m_1y + n_1 = 0$ and $L_2: l_2x + m_2y + n_2 = 0$.The distanc

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$f^4 - g^4 = c(bf^2 - ag^2)$

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(C) The given equation represents a pair of straight lines. Let the lines be $L_1: l_1x + m_1y + n_1 = 0$ and $L_2: l_2x + m_2y + n_2 = 0$.The distance of these lines from the origin $(0,0)$ is given by $d_1 = \frac{|n_1|}{\sqrt{l_1^2 + m_1^2}}$ and $d_2 = \frac{|n_2|}{\sqrt{l_2^2 + m_2^2}}$.Given $d_1 = d_2$,we have $\frac{n_1^2}{l_1^2 + m_1^2} = \frac{n_2^2}{l_2^2 + m_2^2}$.Comparing the given equation with the product of the two lines $(l_1x + m_1y + n_1)(l_2x + m_2y + n_2) = 0$,we have $l_1l_2 = a$,$m_1m_2 = b$,$n_1n_2 = c$,$l_1n_2 + l_2n_1 = 2g$,and $m_1n_2 + m_2n_1 = 2f$.From the condition $\frac{n_1^2}{l_1^2 + m_1^2} = \frac{n_2^2}{l_2^2 + m_2^2}$,we can derive $n_1^2(l_2^2 + m_2^2) = n_2^2(l_1^2 + m_1^2)$.Using the relations between coefficients,this simplifies to $f^4 - g^4 = c(bf^2 - ag^2)$.

The lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ will be equidistant from the origin,if

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