Pair of Linear Equations in Two Variables | Exercise 3.1

Question 6

Given the linear equation $2x + 3y - 8 = 0$ , write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

Solution

Understand the Question

For a pair of linear equations $a_1x + b_1y + c_1 = 0$ $a_{1} x + b_{1} y + c_{1} = 0$ and $a_2x + b_2y + c_2 = 0$ $a_{2} x + b_{2} y + c_{2} = 0$ , the geometric relationship depends on the ratios of their coefficients:
- Intersecting lines: $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$
- Parallel lines: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$
- Coincident lines: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$
Given equation: $2x + 3y - 8 = 0$ , where $a_1 = 2$ , $b_1 = 3$ , and $c_1 = -8$ .

(i) intersecting lines

Step 1 · Find an Equation for Intersecting Lines

Diagram 1

For intersecting lines, the condition is: $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

Let $a_2 = 1$ , $b_2 = 1$ , and $c_2 = 1$ .

Checking the ratios:

\begin{aligned} \dfrac{a_1}{a_2} &= \dfrac{2}{1} = 2 \\[0.6em] \dfrac{b_1}{b_2} &= \dfrac{3}{1} = 3 \end{aligned}

Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ , the condition is satisfied.

Therefore, a possible second equation is: $x + y + 1 = 0$

Answer

(i) $x + y + 1 = 0$

(ii) parallel lines

Step 1 · Find an Equation for Parallel Lines

Diagram 2

For parallel lines, the condition is: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

Choose $a_2 = 4$ and $b_2 = 6$ , giving: $\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}$

We require $\dfrac{c_1}{c_2} \neq \dfrac{1}{2} \implies \dfrac{-8}{c_2} \neq \dfrac{1}{2} \implies c_2 \neq -16$ .

Choosing $c_2 = 1$ gives: $\dfrac{c_1}{c_2} = \dfrac{-8}{1} = -8 \neq \dfrac{1}{2}$

Therefore, a possible second equation is: $4x + 6y + 1 = 0$

Answer

(ii) $4x + 6y + 1 = 0$

(iii) coincident lines

Step 1 · Find an Equation for Coincident Lines

Diagram 3

For coincident lines, the condition is: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

Multiplying the entire given equation by $2$ :

\begin{aligned} 2(2x + 3y - 8) &= 2(0) \\[0.6em] 4x + 6y - 16 &= 0 \end{aligned}

Here, $a_2 = 4$ , $b_2 = 6$ , and $c_2 = -16$ .

Checking the ratios: $\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-16} = \dfrac{1}{2}$

Since all ratios are equal, the condition is satisfied.

Therefore, a possible second equation is: $4x + 6y - 16 = 0$

Answer

(iii) $4x + 6y - 16 = 0$

Common Mistakes

Multiple Valid Answers: These questions have infinitely many valid equations. Any equation that satisfies the ratio condition is correct.
Parallel vs. Coincident: For parallel lines, ensuring $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$ is not enough; you must also verify that $\dfrac{c_1}{c_2}$ is different, otherwise the lines become coincident.

Question 6

More questions in Exercise 3.1