Notes On Angle Between two Lines - CBSE Class 11 Maths
Consider two lines L1 and L2 in a coordinate plane with inclinations a1 and a2. If α1 = α2 ⇒ l1 || l2 If α1 ≠ α2 ⇒ l1 and l2 are intersecting lines The intersecting lines L1 and L2 form two pairs of vertically opposite equal angles. α1, α2 ≠ 90o Consider triangle ABC. In ∆ABC: ∠ABX = ∠BCA + ∠BAC (Exterior angle = Sum of interior opposite angles) ⇒ α2 = α1 + θ Or θ = α2 - α1 Thus, tan θ = tan (α2 - α1) Or Tan θ = (Tan α2 - Tan α1)/1 + Tan α1 x Tan α2 …..(1) Tan α2 = m2 (Slope of line l2) Tan α1 = m1 (Slope of line l1) Thus: Tan θ = (m2 - m1)/1 + m1m2 ….(2) Now θ + Φ = 180o (Supplementary angles) ⇒ Φ = 180o - θ Tan Φ = tan (180o - θ) = - tan θ …..(3) ⇒ Tan Φ = - (m2 - m1)/1 + m1m2 ….(4) Tan θ = [(m2 - m1)/1 + m1m2] ….(2) Tan Φ = - [(m2 - m1)/1 + m1m2] ….(4) Case I: [(m2 - m1)/1 + m1m2] is positive ⇒ tan θ is positive ⇒ θ is an acute angle Case I: [(m2 - m1)/1 + m1m2] is positive ⇒ tan Φ is negative ⇒ Φ is an obtuse angle Case II: [(m2 - m1)/1 + m1m2] is negative ⇒ tan θ is negative ⇒ θ is an obtuse angle Case II: [(m2 - m1)/1 + m1m2] is negative ⇒ tan Φ is positive ⇒ Φ is an acute angle tan θ = ⎢(m2 - m1)/1 + m1m2 ⎢ Φ = 180o - θ Collinearity of Points If A, B and C are collinear: Area of ∆ABC = ½ ∣x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)∣ = 0 Collinearity of points by using the slopes of lines passing through them: Consider two lines AB and BC passing through the given points. Let m1 and m2 be the slopes of lines AB and BC, respectively. If A, B and C are collinear: ∠ABC (θ) = 180o tan θ = (m2 - m1)/1 + m1m2 ….(1) If θ = 180o, tan θ = 0 ⇒ (m2 - m1)/1 + m1m2 = 0 Or m2 - m1 = 0 Or m2 = m1 If three given points are collinear, then the slopes of the lines passing through any two of them are equal. If the slopes of two lines passing through any two of three given points are equal, the given points are collinear. A(2, -1), B(6, 4), C(10, 9) Slope of AB = (y2 - y1)/(x2 - x1) = {4 - (-1)}/(6 - 2)= 5/4 Slope of BC = (9-5)/(10-6) = 5/4 Slope of AB = Slope of BC = 5/4 Slope of AC = [9 - (-1)]/(10 - 2)= 10/8 = 5/4 Slope of AB = Slope of BC = Slope of AC = 5/4.

#### Summary

Consider two lines L1 and L2 in a coordinate plane with inclinations a1 and a2. If α1 = α2 ⇒ l1 || l2 If α1 ≠ α2 ⇒ l1 and l2 are intersecting lines The intersecting lines L1 and L2 form two pairs of vertically opposite equal angles. α1, α2 ≠ 90o Consider triangle ABC. In ∆ABC: ∠ABX = ∠BCA + ∠BAC (Exterior angle = Sum of interior opposite angles) ⇒ α2 = α1 + θ Or θ = α2 - α1 Thus, tan θ = tan (α2 - α1) Or Tan θ = (Tan α2 - Tan α1)/1 + Tan α1 x Tan α2 …..(1) Tan α2 = m2 (Slope of line l2) Tan α1 = m1 (Slope of line l1) Thus: Tan θ = (m2 - m1)/1 + m1m2 ….(2) Now θ + Φ = 180o (Supplementary angles) ⇒ Φ = 180o - θ Tan Φ = tan (180o - θ) = - tan θ …..(3) ⇒ Tan Φ = - (m2 - m1)/1 + m1m2 ….(4) Tan θ = [(m2 - m1)/1 + m1m2] ….(2) Tan Φ = - [(m2 - m1)/1 + m1m2] ….(4) Case I: [(m2 - m1)/1 + m1m2] is positive ⇒ tan θ is positive ⇒ θ is an acute angle Case I: [(m2 - m1)/1 + m1m2] is positive ⇒ tan Φ is negative ⇒ Φ is an obtuse angle Case II: [(m2 - m1)/1 + m1m2] is negative ⇒ tan θ is negative ⇒ θ is an obtuse angle Case II: [(m2 - m1)/1 + m1m2] is negative ⇒ tan Φ is positive ⇒ Φ is an acute angle tan θ = ⎢(m2 - m1)/1 + m1m2 ⎢ Φ = 180o - θ Collinearity of Points If A, B and C are collinear: Area of ∆ABC = ½ ∣x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)∣ = 0 Collinearity of points by using the slopes of lines passing through them: Consider two lines AB and BC passing through the given points. Let m1 and m2 be the slopes of lines AB and BC, respectively. If A, B and C are collinear: ∠ABC (θ) = 180o tan θ = (m2 - m1)/1 + m1m2 ….(1) If θ = 180o, tan θ = 0 ⇒ (m2 - m1)/1 + m1m2 = 0 Or m2 - m1 = 0 Or m2 = m1 If three given points are collinear, then the slopes of the lines passing through any two of them are equal. If the slopes of two lines passing through any two of three given points are equal, the given points are collinear. A(2, -1), B(6, 4), C(10, 9) Slope of AB = (y2 - y1)/(x2 - x1) = {4 - (-1)}/(6 - 2)= 5/4 Slope of BC = (9-5)/(10-6) = 5/4 Slope of AB = Slope of BC = 5/4 Slope of AC = [9 - (-1)]/(10 - 2)= 10/8 = 5/4 Slope of AB = Slope of BC = Slope of AC = 5/4.

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