The product of the perpendiculars drawn from | vedclass

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(D) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Let the two lines be $L_1: l_1x + m_1y + n_1 = 0$ and

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$\frac{c}{\sqrt{(a - b)^2 + 4h^2}}$

Vedclass · Answer

(D) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Let the two lines be $L_1: l_1x + m_1y + n_1 = 0$ and $L_2: l_2x + m_2y + n_2 = 0$.The product of the perpendiculars from the origin $(0, 0)$ to these lines is given by the formula $\left| \frac{n_1}{\sqrt{l_1^2 + m_1^2}} ight| 	imes \left| \frac{n_2}{\sqrt{l_2^2 + m_2^2}} ight|$.For the given equation,the product of the perpendiculars from the origin is $\frac{|c|}{\sqrt{(a-b)^2 + 4h^2}}$.

The product of the perpendiculars drawn from the origin to the lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is:

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