Example 15 - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at Feb. 3, 2020 by Teachoo
Last updated at Feb. 3, 2020 by Teachoo
Transcript
Example 15 Find the angle between the lines y − √3x − 5 = 0 and √3y − x + 6 = 0 . Let the lines be y − √3 x − 5 = 0 √3 y − x + 6 = 0 We know that angle between 2 lines (θ) can be found by using formula tan θ =|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Let the slope of line (1) be m1 & slope of line (2) be m2 Calculating m1 From (1) y – √3x – 5 = 0 y = √3x + 5 The above equation is of the form y = mx + c where m is the slope Thus, m1 = √𝟑 Calculating m1 From (2) √3 y – x + 6 = 0 √3 y = x − 6 y = (𝑥 − 6)/√3 y = 1/√3x − 6/√3 The above equation is of the form y = mx + c where m is the slope Thus, m2 = 𝟏/√𝟑 Angle between two lines is given tan θ =|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Putting values tan θ = |(1/√3 − √3)/(1 + 1/√3 × √3)| = |(1/√3 − √3)/(1 + 1)| = |((1 − (√3)(√3))/√3)/2| = |(1 − 3)/(2√3)| = |( −2)/(2√3)| = |( −1)/√3| = 1/√3 Thus, tan θ = 1/√3 We know that tan 30° = 1/√3 So, tan θ = tan 30° θ = 30° Thus, the acute angle between the lines (1) & (2) is θ = 30° & obtuse angle between (say ϕ) these two lines is ϕ = 180 − θ ϕ = 180 − 30° ϕ = 150° Thus, the required angle between lines is 30° or 150° Note There are always two angles between the lines, one acute angle θ & other obtuse angle 𝝓 which are in linear pair, Thus θ + ϕ = 180° ϕ = 180° – θ
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