Question 2 – Comparing ratios to determine nature of lines
📑 Question Parts
📚 Important Concept
For a pair of linear equations:
\[ a_1x + b_1y + c_1 = 0 \] \[ a_2x + b_2y + c_2 = 0 \]Three cases arise:
- Case 1: If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) → Lines intersect at one point (Unique solution, Consistent)
- Case 2: If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) → Lines are coincident (Infinitely many solutions, Dependent & Consistent)
- Case 3: If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) → Lines are parallel (No solution, Inconsistent)
Part (i)
\( 5x – 4y + 8 = 0 \)
\( 7x + 6y – 9 = 0 \)
Solution Steps
For equation \( 5x – 4y + 8 = 0 \):
\( a_1 = 5 \), \( b_1 = -4 \), \( c_1 = 8 \)
For equation \( 7x + 6y – 9 = 0 \):
\( a_2 = 7 \), \( b_2 = 6 \), \( c_2 = -9 \)
Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{5}{7} \]Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3} \]Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{8}{-9} = \frac{-8}{9} \]We have:
\[ \frac{a_1}{a_2} = \frac{5}{7} \] \[ \frac{b_1}{b_2} = \frac{-2}{3} \]To compare \( \frac{5}{7} \) and \( \frac{-2}{3} \):
\( \frac{5}{7} \approx 0.714 \) and \( \frac{-2}{3} \approx -0.667 \)
Clearly, \( \frac{5}{7} \neq \frac{-2}{3} \)
Therefore: \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at exactly one point.
The pair of equations has a unique solution and is consistent.
Part (ii)
\( 18x + 6y + 24 = 0 \)
Solution Steps
For equation \( 9x + 3y + 12 = 0 \):
\( a_1 = 9 \), \( b_1 = 3 \), \( c_1 = 12 \)
For equation \( 18x + 6y + 24 = 0 \):
\( a_2 = 18 \), \( b_2 = 6 \), \( c_2 = 24 \)
Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2} \]Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \]Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2} \]We observe that:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2} \]All three ratios are equal!
If we multiply the first equation by 2:
\[ 2(9x + 3y + 12) = 0 \] \[ 18x + 6y + 24 = 0 \]This is exactly the second equation! This confirms that both equations represent the same line.
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident.
The pair of equations has infinitely many solutions and is dependent (consistent).
Part (iii)
\( 2x – y + 9 = 0 \)
Solution Steps
For equation \( 6x – 3y + 10 = 0 \):
\( a_1 = 6 \), \( b_1 = -3 \), \( c_1 = 10 \)
For equation \( 2x – y + 9 = 0 \):
\( a_2 = 2 \), \( b_2 = -1 \), \( c_2 = 9 \)
Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{6}{2} = 3 \]Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{-3}{-1} = 3 \]Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{10}{9} \]We observe that:
\[ \frac{a_1}{a_2} = 3 \] \[ \frac{b_1}{b_2} = 3 \] \[ \frac{c_1}{c_2} = \frac{10}{9} \approx 1.111 \]Therefore: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
(Since \( 3 \neq \frac{10}{9} \))
Rewrite both equations in slope-intercept form \( y = mx + c \):
From \( 6x – 3y + 10 = 0 \): \( y = 2x + \frac{10}{3} \) → slope = 2
From \( 2x – y + 9 = 0 \): \( y = 2x + 9 \) → slope = 2
Both lines have the same slope but different y-intercepts, confirming they are parallel.
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel.
The pair of equations has no solution and is inconsistent.
📊 Summary of All Three Parts
| Part | Equations | \( \frac{a_1}{a_2} \) | \( \frac{b_1}{b_2} \) | \( \frac{c_1}{c_2} \) | Condition | Nature of Lines | Solution Type |
|---|---|---|---|---|---|---|---|
| (i) | 5x – 4y + 8 = 0 7x + 6y – 9 = 0 | \( \frac{5}{7} \) | \( \frac{-2}{3} \) | \( \frac{-8}{9} \) | \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) | Intersecting | Unique solution (Consistent) |
| (ii) | 9x + 3y + 12 = 0 18x + 6y + 24 = 0 | \( \frac{1}{2} \) | \( \frac{1}{2} \) | \( \frac{1}{2} \) | \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) | Coincident | Infinitely many solutions (Dependent) |
| (iii) | 6x – 3y + 10 = 0 2x – y + 9 = 0 | 3 | 3 | \( \frac{10}{9} \) | \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) | Parallel | No solution (Inconsistent) |
⚠️ Common Mistakes to Avoid
✅ Correct: Always include the sign. For example, if \( b_1 = -4 \), then \( \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3} \), not \( \frac{2}{3} \).
✅ Correct: Remember: Parallel means \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Coincident means all three ratios are equal.
✅ Correct: Always simplify ratios. For example, \( \frac{9}{18} = \frac{1}{2} \) and \( \frac{3}{6} = \frac{1}{2} \) makes comparison easier.
✅ Correct: You must calculate and compare all three ratios \( \frac{a_1}{a_2} \), \( \frac{b_1}{b_2} \), and \( \frac{c_1}{c_2} \) to determine the nature of lines correctly.
💡 Key Points to Remember
- Intersecting Lines: \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) → Unique solution (Consistent)
- Coincident Lines: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) → Infinitely many solutions (Dependent & Consistent)
- Parallel Lines: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) → No solution (Inconsistent)
- Standard Form: Always write equations in the form \( ax + by + c = 0 \) before identifying coefficients.
- Sign Matters: Include negative signs when calculating ratios of coefficients.
- Simplify Ratios: Reduce fractions to lowest terms for easier comparison.
- Graphical Interpretation: This algebraic method tells us about the graphical relationship without actually drawing the graphs.
- Quick Check: For coincident lines, one equation should be a multiple of the other.
📝 Practice Similar Problems
Farhan Mansuri
M.Sc. Mathematics, B.Ed.
15+ Years Teaching Experience
Passionate about making mathematics easy and enjoyable for students.