Question

If pair of linear equations given by \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) has infinitely many solutions, then which of the following is definitely true ?

• A. \(\frac{a_1}{a_2} = \frac{c_2}{c_1}\)
• B. \(a_1a_2 \neq b_1b_2\)
• C. \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
• D. \(a_1b_2 = a_2b_1\)

Hint:

Remember the three algebraic conditions for a pair of linear equations:
1. Unique solution: \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
2. No solution: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
3. Infinitely many solutions: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)

Answer 1

The Correct Option is D

Step 1: Recall the condition.
Infinitely many solutions means the two lines are coincident, so $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Step 2: Test the condition with a concrete coincident pair instead of pure algebra.
Take $x+2y+3=0$ and $2x+4y+6=0$ (the second is just the first doubled, so they are coincident). Here $a_1=1,b_1=2,c_1=3,a_2=2,b_2=4,c_2=6$.
Step 3: Check each option numerically.
$a_1b_2 = 1 \times 4 = 4$ and $a_2b_1 = 2 \times 2 = 4$, so $a_1b_2 = a_2b_1$ holds. Checking the other options with these numbers fails to hold in general, so option (D) is the one that is always true.
\[ \boxed{a_1b_2 = a_2b_1} \]