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

For any pair of linear equations,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

(i) For intersecting lines

Condition : $\dfrac{a$1}{a2} ≠ $\dfrac{b$1}{b2}.

For,

2x + 3y - 8 = 0

a₁ = 2 and b₁ = 3

So, considering a₂ = 3 and b₂ = 2 will satisfy the condition for intersecting lines. c₂ can be any value.

Let c₂ be -7.

So,

⇒ a₂x + b₂y + c₂ = 0

⇒ 3x + 2y - 7 = 0.

Hence, equation of required line is 3x + 2y - 7 = 0.

(ii) For parallel lines

Condition : $\dfrac{a$1}{a2} = $\dfrac{b$1}{b2} ≠ $\dfrac{c$1}{c2}.

For,

2x + 3y - 8 = 0

a₁ = 2, b₁ = 3 and c₁ = -8.

So, considering a₂ = 2, b₂ = 3 and c₂ = -12 will satisfy the condition for parallel lines.

$\dfrac{a$1}{a2} = \dfrac{2}{2} = 1

$\dfrac{b$1}{b2} = \dfrac{3}{3} = 1

$\dfrac{c$1}{c2} = \dfrac{-8}{-12} = \dfrac{2}{3}

Since, $\dfrac{a$1}{a2} = $\dfrac{b$1}{b2} ≠ $\dfrac{c$1}{c2}.

Substituting values we get :,

⇒ a₂x + b₂y + c₂ = 0

⇒ 2x + 3y - 12 = 0.

Hence, equation of required line is 2x + 3y - 12 = 0.

(iii) For coincident lines

Condition : $\dfrac{a$1}{a2} = \dfrac{b1}{b2} = \dfrac{c1}{c2}.

For,

2x + 3y - 8 = 0

a₁ = 2, b₁ = 3 and c₁ = -8.

So, considering a₂ = 4, b₂ = 6 and c₂ = -16 will satisfy the condition for coincident lines.

$\dfrac{a$1}{a2} = \dfrac{2}{4} = \dfrac{1}{2}

$\dfrac{b$1}{b2} = \dfrac{3}{6} = \dfrac{1}{2}

$\dfrac{c$1}{c2} = \dfrac{-8}{-16} = \dfrac{1}{2}

Since, $\dfrac{a$1}{a2} = \dfrac{b1}{b2} = \dfrac{c1}{c2}.

Substituting values we get :,

⇒ a₂x + b₂y + c₂ = 0

⇒ 4x + 6y - 16 = 0.

Hence, equation of required line is 4x + 6y - 16 = 0.