Question 6

We will use the conditions for ratios of coefficients to find the second linear equation.

Step 1 — Intersecting Lines

Let the given equation be $L_1: 2x + 3y - 8 = 0$. Here, $a_1 = \mathbf{2}$, $b_1 = \mathbf{3}$, $c_1 = \mathbf{-8}$. For intersecting lines, the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. Let's choose $a_2 = \mathbf{1}$ and $b_2 = \mathbf{1}$. We check the ratio of coefficients.

$\frac{a_1}{a_2} = \frac{2}{1}$

$= 2$

$\frac{b_1}{b_2} = \frac{3}{1}$

$= 3$

Since $2 \neq 3$, the condition is satisfied. We can choose any value for $c_2$. Let's choose $c_2 = \mathbf{1}$. So, the second equation is $1x + 1y + 1 = 0$.

$\boxed{x + y + 1 = 0}$

Step 2 — Parallel Lines

Let the given equation be $L_1: 2x + 3y - 8 = 0$. Here, $a_1 = \mathbf{2}$, $b_1 = \mathbf{3}$, $c_1 = \mathbf{-8}$. For parallel lines, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Let's choose $a_2$ and $b_2$ such that $\frac{a_1}{a_2} = \frac{b_1}{b_2}$. We can multiply $a_1$ and $b_1$ by a common factor, say 2. So, $a_2 = 2 \times 2 = \mathbf{4}$ and $b_2 = 2 \times 3 = \mathbf{6}$. Now, we need to choose $c_2$ such that $\frac{c_1}{c_2}$ is not equal to $\frac{a_1}{a_2}$. The ratio $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$. So, we need $\frac{-8}{c_2} \neq \frac{1}{2}$. This means $c_2 \neq -16$. Let's choose $c_2 = \mathbf{1}$. The second equation is $4x + 6y + 1 = 0$.

$\boxed{4x + 6y + 1 = 0}$

Step 3 — Coincident Lines

Let the given equation be $L_1: 2x + 3y - 8 = 0$. Here, $a_1 = \mathbf{2}$, $b_1 = \mathbf{3}$, $c_1 = \mathbf{-8}$. For coincident lines, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. This means the second equation is a multiple of the first equation. Let's multiply the entire first equation by a constant, say 2.

$2 \times (2x + 3y - 8) = 2 \times 0$

$4x + 6y - 16 = 0$

Here, $a_2 = \mathbf{4}$, $b_2 = \mathbf{6}$, $c_2 = \mathbf{-16}$. Let's check the ratios. $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$. $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$. $\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$. All ratios are equal, so the condition is satisfied. The second equation is $4x + 6y - 16 = 0$.

$\boxed{4x + 6y - 16 = 0}$

Answer

(i) For intersecting lines: $x + y + 1 = 0$ (ii) For parallel lines: $4x + 6y + 1 = 0$ (iii) For coincident lines: $4x + 6y - 16 = 0$