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Question
The slope of a line is double of the slope of another line. If tangent of the angle between them is `1/3`, find the slopes of the lines.
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Solution
Let m1 and m be the slopes of the two given lines such that `m_1 =2m`
We know that if θ is the angle between the lines l1 and l2 with slopes m1 and m2, then
`tan theta = |(m_2 - m_1)/(1 + m_1m_2)|`
It is given that the tangent of the angle between the two lines is `1/3`.
`1/3 = |(m - 2m)/(1 + (2m).m)|`
`1/3 = |(- m)/(1 + 2m^2)|`
`1/3 = (- m)/(1 + 2m^2) or 1/3 = -((-m)/(1 + 2m^2)) = m/(1 + 2m^2)`
Case I
= `1/3 = (-m)/(1 + 2m^2)`
= 1 + 2m2 = -3m
= 2m2 + 3m + 1 = 0
= 2m2 + 2m + m + 1 = 0
= 2m(m + 1) +1(m + 1) = 0
= (m + 1) (2m + 1) = 0
= m = -1 or m = `-1/2`
If m = -1, then the slopes of the lines are -1 and -2.
If m =`1/2`, then the slopes of the lines are `1/2` and -1.
Case II
= `1/3 = m/(1 + 2m^2)`
= 2m2 + 1= 3m
= 2m2 - 3m + 1 = 0
= 2m2 - 2m - m + 1 = 0
= 2m(m - 1) +1(m - 1) = 0
= (m - 1) (2m - 1) = 0
= m = 1 or m = `1/2`
If m = 1, then the slopes of the lines are 1 and 2.
If m = `1/2`, then the slopes of the lines are `1/2` and 1.
Hence, the slopes of the lines are -1 and -2 or `-1/2` and -1 or 1 and 2 or `1/2 and1.`
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