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
The given equation is
\[ 2x + 3y - 8 = 0. \]
Its coefficients are \(a_1 = 2, b_1 = 3, c_1 = -8\).
(i) Intersecting lines
For intersecting lines, we need
\[ \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}. \]
Take another equation, for example
\[ 3x + 2y - 7 = 0. \]
Here \(a_2 = 3, b_2 = 2\). Then
\[ \dfrac{a_1}{a_2} = \dfrac{2}{3}, \quad \dfrac{b_1}{b_2} = \dfrac{3}{2}. \]
Since these ratios are unequal, the two lines intersect at a unique point.
(ii) Parallel lines
For parallel but distinct lines, we require
\[ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}. \]
Choose an equation with proportional \(a\) and \(b\) but different constant term. Multiply the whole given equation by 2 and change the constant term:
\[ 4x + 6y - 16 = 0 \quad \text{(same ratios as original)}. \]
Now \(a_2 = 4, b_2 = 6, c_2 = -16\). We have
\[ \dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-16} = \dfrac{1}{2}. \]
To make them distinct, keep \(a_2, b_2\) proportional but alter \(c_2\). One convenient choice shown in the textbook is
\[ 2x + 3y - 12 = 0. \]
Here
\[ \dfrac{a_1}{a_2} = \dfrac{2}{2} = 1, \quad \dfrac{b_1}{b_2} = \dfrac{3}{3} = 1, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-12} = \dfrac{2}{3}. \]
Thus \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\), so the two lines are parallel and distinct.
(iii) Coincident lines
For coincident lines, all three ratios must be equal:
\[ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}. \]
Take an equation which is a non-zero multiple of the given equation, for example multiply by 2:
\[ 4x + 6y - 16 = 0. \]
Now
\[ \dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-16} = \dfrac{1}{2}. \]
All three are equal, so the two equations represent the same line and are coincident.
A set of acceptable answers is therefore:
(i) \(3x + 2y - 7 = 0\) (intersecting with the given line),
(ii) \(2x + 3y - 12 = 0\) (parallel to the given line),
(iii) \(4x + 6y - 16 = 0\) (coincident with the given line).
Use coefficient ratios: change them so \(\tfrac{a_1}{a_2}\) and \(\tfrac{b_1}{b_2}\) differ for intersecting lines, keep them equal but alter the constant for parallel lines, and multiply the whole equation by a constant for coincident lines.