Solution: For any pair of linear equation, a₁ x + b₁ y + c₁ = 0 a₂ x + b₂ y + c₂ = 0 a) If a₁/a₂ ≠ b₁/b₂ (Intersecting Lines) b) If a₁/a₂ = b₁/b₂ = c₁/c₂ (Coincident Lines) c) If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (Parallel Lines) (i) Intersecting lines Condition: a₁/a₂ ≠ b₁/b₂ 2x + 3y - 8 = 0 a₁ = 2 b₁ = 3 So, considering a₂ = 3 and b₂ = 2 will satisfy the condition for intersecting lines. c₂ can be any value. a₁/a₂ = 2/3 b₁/b₂ = 3/2 2/3 ≠ 3/2 Therefore, another linear equation is 3x + 2y - 6 = 0 (ii) Parallel lines Condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ 2x + 3y - 8 = 0 a₁ = 2 b₁ = 3 c₁ = - 8 So, considering a₂ = 4, b₂ = 6, c₂ = 9 will satisfy the condition for parallel lines. a₁/a₂ = 2/4 = 1/2 b₁/b₂ = 3/6 = 1/2 c₁/c₂ = - 8/9 Thus, a₁/a₂ = b₁/b₂ ≠ c₁/c₂ Therefore, another linear equation is 4x + 6y + 9 = 0 (iii) Coincident lines Condition: a₁/a₂ = b₁/b₂ = c₁/c₂ 2x + 3y - 8 = 0 Condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ 2x + 3y - 8 = 0 We know that, a₁= 2, b₁= 3, c₁= - 8 So, considering a₂ = 4, b₂ = 6, c₂ = - 16 will satisfy the condition for coincident lines. a₁/a₂ = 2/4 = 1/2 b₁/b₂ = 3/6 = 1/2 c₁/c₂ = - 8/(-16) = 1/2 Thus, a₁/a₂ = b₁/b₂ = c₁/c₂ Therefore, linear equation is 4x + 6y -16 = 0 ☛ Check: Class 10 Maths NCERT Solutions Chapter 3 Video Solution: NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.2 Question 6 Summary: Given the linear equation 2x + 3y - 8 = 0, another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines is 3x + 2y - 6 = 0, for parallel lines is 4x + 6y + 9 = 0 and for the coincident lines is 4x + 6y - 16 = 0. ☛ Related Questions:
Ans) 1) intersecting = 4x-3y-4=0 2) coincident = 4x-10y-8=0 3) parallel = 2x-5y-2=0 Hope it helps.
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