$If α, β are inclinations made by the perpendiculars drawn from the origin to two$
$lines L_{1}, L_{2} and P_{1}, P_{2} are the perpendicular distances from the origin to L_{1}, L_{2}$
$respectively. The angle between two lines L_{1}, L_{2} is$

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$β - α$

$α - β$

$90 °$

$β + α$

answer is B.

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Detailed Solution

Normal form of a line:
If $α$ is the inclination made by the perpendicular drawn from the origin to the line $L_{1}$ and the length of the perpendicular drawn from origin to the line $L_{1}$ is $p_{1}$ then the equation of the line $L_{1}$ is $x \cos α + y \sin α = p_{1}$

$Similarly, the equation of the line L_{2} is x \cos β + y \sin β = p_{2}$

$Use formula m = - \frac{a}{b} for the line a x + b y + c = 0$

$The slopes of lines L_{1}, L_{2} are m_{1} = - \frac{1}{\tan α} and m_{2} = - \frac{1}{\tan β},$

$Suppose that θ be the angle between two lines L_{1}, L_{2} then$

$\begin{matrix} \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| \\ = |\frac{- \frac{1}{\tan α} - \frac{1}{\tan β}}{1 + (- \frac{1}{\tan α}) (- \frac{1}{\tan β})}| \\ = |\frac{\tan β - \tan α}{1 + \tan β \tan α}| \\ = \tan (β - α) \end{matrix}$