Question:

1. Given line : \(2x+3y-8=0\) \(\Rightarrow a_1=2,\; b_1=3,\; c_1=-8\).

2. Choose a second line \(a_2x+b_2y+c_2=0\) such that the ratios of the coefficients satisfy the required condition.

(i) Intersecting lines
- Choose coefficients so that \(\frac{a_1}{a_2}
eq\frac{b_1}{b_2}\).
- Example: take \(a_2=3,\; b_2=-2\). Then \(\frac{2}{3}
eq\frac{3}{-2}\).
- Choose any constant term, say \(c_2=5\).
- Hence the second equation is \(3x-2y+5=0\).
- Since the ratios of \(x\) and \(y\) coefficients are different, the two lines intersect at a unique point.

(ii) Parallel lines
- For parallelism we need \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) but \(\frac{c_1}{c_2}\) must be different.
- Multiply the given coefficients by a non‑zero constant, e.g. \(k=2\): \(a_2=2\times2=4,\; b_2=2\times3=6\).
- Keep the same ratio for \(c\) different from \(k\). Let \(c_2=-12\) (instead of \(-16\)).
- Second equation: \(4x+6y-12=0\).
- Here \(\frac{2}{4}=\frac{3}{6}=\frac{1}{2}\) but \(\frac{-8}{-12}=\frac{2}{3}
eq\frac{1}{2}\); therefore the lines are distinct and parallel.

(iii) Coincident lines
- For coincidence the three ratios must be equal.
- Use the same constant multiple for all coefficients, e.g. \(k=2\).
- Then \(a_2=4,\; b_2=6,\; c_2=-16\).
- Second equation: \(4x+6y-16=0\).
- Since \(\frac{2}{4}=\frac{3}{6}=\frac{-8}{-16}=\frac{1}{2}\), the two equations represent the same straight line; they are coincident.

3. Verification (optional):
- For (i) solving \(2x+3y-8=0\) and \(3x-2y+5=0\) simultaneously gives a unique solution, confirming intersection.
- For (ii) the slopes are \(-\frac{2}{3}\) for both lines, confirming parallelism.
- For (iii) the second equation is exactly twice the first, confirming coincidence.