Let us consider ‘m_{1}’ and ‘m’ be the slope of the two given lines such that m_{1 }= 2m

We know that if θ is the angle between the lines l_{1} and l_{2} with slope m_{1} and m_{2}, then

\( |\dfrac{m_2+m_1}{1+m_1m_2}|\)

It is also given that the tangent of the angle between the two lines is\( \dfrac{1}{3} \)

\( \dfrac{1}{3}=|\dfrac{m-2m}{1+(2m)\times m}| \)

= \( \dfrac{1}{3}=|\dfrac{-m}{1+2m^2}| \) or

= \( |\dfrac{-m}{1+2m^2}|\)

Case 1:-

\( \dfrac{1}{3}=|\dfrac{-m}{1+2m^2}| \)

1+2m^{2} = -3m

2m^{2} +1 +3m = 0

2m (m+1) + 1(m+1) = 0

(2m+1) (m+1)= 0

m = -1 or \(- \dfrac{1}{2} \)

If m = -1, then the slope of the lines are -1 and -2

If m = \( - \dfrac{1}{2} \), then the slope of the lines are \( - \dfrac{1}{2} \) and -1

Case 2:

\( \dfrac{1}{3}=|\dfrac{-m}{1+2m^2}| \)

2m^{2} – 3m + 1 = 0

2m^{2} – 2m – m + 1 = 0

2m (m – 1) – 1(m – 1) = 0

m = 1 or \( \dfrac{1}{2} \)

If m = 1, then the slope of the lines are 1 and 2

If m = \( \dfrac{1}{2} \), then the slope of the lines are \( \dfrac{1}{2} \) and 1

The slope of the lines are [-1 and -2] or [\(- \dfrac{1}{2} \) and -1]

or [1 and 2] or [\( \dfrac{1}{2} \) and 1]

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