This is a fundamental question from CBSE Class 10 Maths that tests your understanding of how to determine the relationship between two linear equations without actually solving them. By comparing the ratios of coefficients, you can predict whether lines will intersect, run parallel, or overlap completely!
🔍 Understanding the Ratio Comparison Method
When we have two linear equations in the standard form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), we can determine their relationship by comparing three ratios:
- \(\frac{a_1}{a_2}\) – ratio of coefficients of x
- \(\frac{b_1}{b_2}\) – ratio of coefficients of y
- \(\frac{c_1}{c_2}\) – ratio of constant terms
📌 Quick Reference: The Three Cases
Case 1: \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) → Lines intersect at one point (unique solution)
Case 2: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) → Lines are parallel (no solution)
Case 3: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) → Lines are coincident (infinitely many solutions)
📝 Solving Part (i): 5x – 4y + 8 = 0 and 7x + 6y – 9 = 0
Let’s identify the coefficients first:
For equation 5x – 4y + 8 = 0:
\(a_1 = 5\), \(b_1 = -4\), \(c_1 = 8\)
For equation 7x + 6y – 9 = 0:
\(a_2 = 7\), \(b_2 = 6\), \(c_2 = -9\)
Now let’s calculate the three ratios:
\[ \frac{a_1}{a_2} = \frac{5}{7} \] \[ \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3} \] \[ \frac{c_1}{c_2} = \frac{8}{-9} = \frac{-8}{9} \]
Comparing these ratios: \(\frac{5}{7} \neq \frac{-2}{3}\)
✅ Conclusion: Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines representing these equations intersect at exactly one point. This means the pair of equations has a unique solution.
📝 Solving Part (ii): 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
Let’s identify the coefficients:
For equation 9x + 3y + 12 = 0:
\(a_1 = 9\), \(b_1 = 3\), \(c_1 = 12\)
For equation 18x + 6y + 24 = 0:
\(a_2 = 18\), \(b_2 = 6\), \(c_2 = 24\)
Calculating the ratios:
\[ \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2} \] \[ \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \] \[ \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2} \]Notice something interesting? All three ratios are equal!
✅ Conclusion: Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}\), the lines are coincident (they overlap completely). This means the second equation is just a multiple of the first – if you multiply the first equation by 2, you get the second equation! The system has infinitely many solutions.
📝 Solving Part (iii): 6x – 3y + 10 = 0 and 2x – y + 9 = 0
Identifying the coefficients:
For equation 6x – 3y + 10 = 0:
\(a_1 = 6\), \(b_1 = -3\), \(c_1 = 10\)
For equation 2x – y + 9 = 0:
\(a_2 = 2\), \(b_2 = -1\), \(c_2 = 9\)
Calculating the ratios:
\[ \frac{a_1}{a_2} = \frac{6}{2} = 3 \] \[ \frac{b_1}{b_2} = \frac{-3}{-1} = 3 \] \[ \frac{c_1}{c_2} = \frac{10}{9} \]Here we can see that the first two ratios are equal, but the third one is different.
✅ Conclusion: Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = 3\) but \(\frac{c_1}{c_2} = \frac{10}{9} \neq 3\), the lines are parallel. They have the same slope but different y-intercepts, so they never meet. The system has no solution.
📊 Comparison Table: All Three Parts
| Part | Equations | \(\frac{a_1}{a_2}\) | \(\frac{b_1}{b_2}\) | \(\frac{c_1}{c_2}\) | Relationship |
|---|---|---|---|---|---|
| (i) | 5x – 4y + 8 = 0 7x + 6y – 9 = 0 |
\(\frac{5}{7}\) | \(\frac{-2}{3}\) | \(\frac{-8}{9}\) | Intersecting (unique solution) |
| (ii) | 9x + 3y + 12 = 0 18x + 6y + 24 = 0 |
\(\frac{1}{2}\) | \(\frac{1}{2}\) | \(\frac{1}{2}\) | Coincident (infinitely many solutions) |
| (iii) | 6x – 3y + 10 = 0 2x – y + 9 = 0 |
\(3\) | \(3\) | \(\frac{10}{9}\) | Parallel (no solution) |
✍️ How to Write This Answer in Your Exam
For Part (i):
Given equations: 5x – 4y + 8 = 0 and 7x + 6y – 9 = 0
Here, \(a_1 = 5\), \(b_1 = -4\), \(c_1 = 8\) and \(a_2 = 7\), \(b_2 = 6\), \(c_2 = -9\)
\(\frac{a_1}{a_2} = \frac{5}{7}\), \(\frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines intersect at one point.
For Part (ii):
Given equations: 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
Here, \(a_1 = 9\), \(b_1 = 3\), \(c_1 = 12\) and \(a_2 = 18\), \(b_2 = 6\), \(c_2 = 24\)
\(\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}\)
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident.
For Part (iii):
Given equations: 6x – 3y + 10 = 0 and 2x – y + 9 = 0
Here, \(a_1 = 6\), \(b_1 = -3\), \(c_1 = 10\) and \(a_2 = 2\), \(b_2 = -1\), \(c_2 = 9\)
\(\frac{a_1}{a_2} = \frac{6}{2} = 3\), \(\frac{b_1}{b_2} = \frac{-3}{-1} = 3\), \(\frac{c_1}{c_2} = \frac{10}{9}\)
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel.
⚠️ Common Mistakes to Avoid
Mistake 1: Forgetting to simplify ratios
Always reduce fractions to their simplest form before comparing. For example, \(\frac{-4}{6}\) should be simplified to \(\frac{-2}{3}\).
Mistake 2: Wrong sign handling
Be very careful with negative signs! In part (i), \(b_1 = -4\) (not +4), and in part (ii), \(c_2 = -9\) (not +9). A wrong sign will give you the wrong answer.
Mistake 3: Confusing parallel and coincident
Remember: Parallel lines have two equal ratios but the third different. Coincident lines have all three ratios equal. Don’t mix these up!
💡 Memory Trick
Think “1-2-3” Rule:
• 1 equal ratio (or none equal) = Lines intersect at 1 point
• 2 equal ratios = Lines are parallel (2 separate lines)
• 3 equal ratios = Lines are coincident (3 letters in “one” – they’re the same line!)
🤔 Frequently Asked Questions
Q1: Do I need to calculate all three ratios every time?
Not always! If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), you can immediately conclude the lines intersect – no need to calculate \(\frac{c_1}{c_2}\). But if the first two ratios are equal, you must check the third ratio to determine if lines are parallel or coincident.
Q2: What if one of the coefficients is zero?
If a coefficient is zero in one equation but not in the other, the ratio will be 0 or undefined, which means the ratios are not equal. The lines will intersect. For example, if \(a_1 = 0\) but \(a_2 \neq 0\), then \(\frac{a_1}{a_2} = 0\), which will be different from other ratios.
Q3: Can I use this method for equations not in standard form?
Yes, but first rearrange them to standard form \(ax + by + c = 0\). For example, if you have \(y = 2x + 3\), rewrite it as \(2x – y + 3 = 0\), then identify \(a = 2\), \(b = -1\), \(c = 3\).
Q4: Why does coincident mean infinitely many solutions?
When lines are coincident, they’re actually the same line drawn twice. Every point on that line satisfies both equations, so there are infinite points (solutions) that work. Think of it like this: if you draw the same line twice, how many intersection points are there? Infinite!
Q5: Is this method faster than solving the equations?
Absolutely! This method takes just 30 seconds to determine the relationship, while actually solving the equations could take 2-3 minutes. In exams where time is precious, this ratio comparison method is a lifesaver. Plus, sometimes you’re only asked about the relationship, not the actual solution!
🔗 Related Questions You Should Practice
- Find the value of k for which the equations 3x + 4y = 12 and kx + 12y = 30 represent parallel lines
- For what value of k will the equations 2x + 3y = 7 and (k+2)x + 6y = 14 have infinitely many solutions?
- Determine graphically whether the pair of equations x + 3y = 6 and 2x – 3y = 12 is consistent
- Check whether the pair of equations 2x + y – 5 = 0 and 3x – 2y – 4 = 0 are consistent
- Solve the pair of linear equations: 3x + 2y = 11 and 2x + 3y = 4
💡 Exam Tip: This type of question is almost guaranteed to appear in your board exam! It’s usually worth 2-3 marks and takes less than 5 minutes if you know the method well. Practice 10-15 similar problems and you’ll be able to solve these in your sleep. The key is accurate calculation of ratios and proper simplification – that’s where most students lose marks!