Ex 10.3, 9 - Class 11

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Ex10.3, 9 Find angles between the lines √3x + y = 1 and x + √3 y = 1 Given equation of lines, √3x + y = 1 x + √3 y = 1 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 slope of line (1) be m1 & slope of line (2) be m2 Now, Angle between lines √3x + y = 1 and x + √3y = 1 is tan θ =|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Putting values tan θ = |( − √3 − (( − 1)/√3))/(1 + ( − √3)(( − 1)/√3) )| tan θ = |( − √3 + 1/√3)/(1 + 1)| tan θ = |((( − √3)(√3) + 1)/√3)/2| tan θ = |( − 3 + 1)/(2√3)| tan θ = |( − 2)/(2√3)| tan θ = |( − 1)/√3| tan θ = 1/√3 We know that tan 30° = 1/√3 tan θ = tan 30° Hence θ = 30° Thus, the acute angle between the lines (1) & (2) is θ = 30° & obtuse angle (ϕ) between these two lines is ϕ = 180 − θ ϕ = 180 − 30° ϕ = 150° Thus, the required angle between lines is 30° or 150°