Coincident lines require 1/(k − 1) = 2/(k + 1) = 3/(k + 3). Cross-multiplying the first pair gives k + 1 = 2k − 2, which solves to k = 3. This value satisfies all ratios equally. NCERT Solution for ...
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Infinitely many solutions require (a + b)/3 = −2b/(−1) = (5a + 2b + 1)/14. Equating gives a = 5b and 5a − 26b = −1. Substituting a = 5b yields b = 1, which gives a = 5. Class 9 ...
For a unique solution, a₁/a₂ ≠ b₁/b₂. Here, 4/2 ≠ p/2, which simplifies to 2 ≠ p/2, giving p ≠ 4. Thus, a unique solution exists for all real values of p except 4. Ganita Manjari Class 9 Maths part 2 ...
Exactly one unique line exists. A specified slope m and a fixed point (x₁, y₁) uniquely determine the line using the point-slope relation y − y₁ = m(x − x₁). Ganita Manjari Part 2 Class 9 Maths chapter 13 question answer Class ...
Infinitely many lines exist with a given slope. In the slope-intercept form y = mx + c, varying the y-intercept c generates an infinite family of parallel lines with the same slope m. Class 9 Maths (Ganita Manjari part two) Chapter ...