Pair of Linear Equations in Two Variables | Exercise 3.1

Question 2

On comparing the ratios $\dfrac{a_1}{a_2}$ , $\dfrac{b_1}{b_2}$ and $\dfrac{c_1}{c_2}$ , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) $5x - 4y + 8 = 0$ $7x + 6y - 9 = 0$

(ii) $9x + 3y + 12 = 0$ $18x + 6y + 24 = 0$

(iii) $6x - 3y + 10 = 0$ $2x - y + 9 = 0$

Solution

Understand the Question

For a pair of linear equations in two variables, $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ , the nature of the lines is determined by comparing the ratio of their coefficients:

Intersecting lines (intersect at exactly one point): $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$
Coincident lines (overlap completely, infinitely many solutions): $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$
Parallel lines (never intersect, no solution): $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

(i) $5x - 4y + 8 = 0$ $7x + 6y - 9 = 0$

Step 1 · Compare Coefficient Ratios

Comparing the given equations with standard forms $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ :

$a_1 = 5, \quad b_1 = -4, \quad c_1 = 8$ $a_2 = 7, \quad b_2 = 6, \quad c_2 = -9$ Diagram 1

Finding the ratios: $\dfrac{a_1}{a_2} = \dfrac{5}{7}$

$\dfrac{b_1}{b_2} = \dfrac{-4}{6} = \dfrac{-2}{3}$

Since $\dfrac{5}{7} \neq \dfrac{-2}{3} \implies \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

Therefore, the lines intersect at a single point.

Answer

(i) Intersect at a point

(ii) $9x + 3y + 12 = 0$ $18x + 6y + 24 = 0$

Step 1 · Compare Coefficient Ratios

Comparing with standard forms:

$a_1 = 9, \quad b_1 = 3, \quad c_1 = 12$ $a_2 = 18, \quad b_2 = 6, \quad c_2 = 24$

Finding the ratios: $\dfrac{a_1}{a_2} = \dfrac{9}{18} = \dfrac{1}{2}$

$\dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}$

$\dfrac{c_1}{c_2} = \dfrac{12}{24} = \dfrac{1}{2}$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} = \dfrac{1}{2}$

Therefore, the lines are coincident.

Answer

(ii) Coincident

(iii) $6x - 3y + 10 = 0$ $2x - y + 9 = 0$

Step 1 · Compare Coefficient Ratios

Comparing with standard forms:

$a_1 = 6, \quad b_1 = -3, \quad c_1 = 10$ $a_2 = 2, \quad b_2 = -1, \quad c_2 = 9$

Finding the ratios: $\dfrac{a_1}{a_2} = \dfrac{6}{2} = 3$

$\dfrac{b_1}{b_2} = \dfrac{-3}{-1} = 3$

$\dfrac{c_1}{c_2} = \dfrac{10}{9}$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

Therefore, the lines are parallel.

Answer

(iii) Parallel

Common Mistakes

Sign Errors: Forgetting negative signs when identifying coefficients (e.g., taking $b_1 = 4$ instead of $-4$ ). Always include the sign preceding each term.
Unnecessary Ratio Calculation: If $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ , you do not need to evaluate $\dfrac{c_1}{c_2}$ ; the lines are confirmed to intersect.
Confusing Parallel and Coincident Conditions: Remember that coincident lines require all three ratios to be equal ( $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ ), whereas parallel lines differ in the constant ratio ( $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$ ).

Question 2

More questions in Exercise 3.1