An equation which can be put in the form ax + by + c = 0 is called a linear equation, where :

(i) x and y are variables,

(ii) a,b and c are real numbers and

(iii) a and b are both not zero.

On drawing a graph for the two linear equations on the same plane, it is seen that only one of the following three posibilities can happen:

(i) the two lines intersect at one point.

(ii) the two lines do not intersect (i.e. the lines are parallel to each other)

(iii) the two lines coincide (i.e. the lines have infinite number of solutions).

On comparing the ratios $\dfrac{a$1}{a2}, \dfrac{b1}{b2},\text{ and } \dfrac{c1}{c2}, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.

(i) 7x - 5y + 10 = 0

6x + 2y - 15 = 0

(ii) 5x + 2y + 8 = 0

15x + 6y + 24 = 0

(iii) 4x - 8y + 9 = 0

2x - 4y + 7 = 0

(iv) x - 2y = 0

3x - 4y - 20 = 0

(v) 2x + 3y - 9 = 0

4x + 6y - 18 = 0

(vi) x + 2y - 4 = 0

2x + 4y - 12 = 0

(i) Given,

Equation 1: 7x - 5y + 10 = 0

a₁ = 7, b₁ = -5, c₁ = 10

Equation 2: 6x + 2y - 15 = 0

a₂ = 6, b₂ = 2, c₂ = -15

$\Rightarrow \dfrac{a$1}{a2} = \dfrac{7}{6}, \dfrac{b1}{b2} = \dfrac{-5}{2}, \dfrac{c1}{c2} = -\dfrac{10}{15}

Since $\dfrac{a$1}{a2} \neq \dfrac{b1}{b2}, the two lines intersect at a point.

Hence, the two lines intersect at a point.

(ii) Given,

Equation 1: 5x + 2y + 8 = 0

a₁ = 5, b₁ = 2, c₁ = 8

Equation 2: 15x + 6y + 24 = 0

a₂ = 15, b₂ = 6, c₂ = 24

$\Rightarrow \dfrac{a$1}{a2} = \dfrac{5}{15} = \dfrac{1}{3}, \dfrac{b1}{b2} = \dfrac{2}{6} = \dfrac{1}{3}, \dfrac{c1}{c2} = \dfrac{8}{24} = \dfrac{1}{3}

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

Hence, the two lines coincide.

(iii) Given,

Equation 1: 4x - 8y + 9 = 0

a₁ = 4, b₁ = -8, c₁ = 9

Equation 2: 2x - 4y + 7 = 0

a₂ = 2, b₂ = -4, c₂ = 7

$\Rightarrow \dfrac{a$1}{a2} = \dfrac{4}{2} = 2, \dfrac{b1}{b2} = \dfrac{-8}{-4} = 2, \dfrac{c1}{c2} = \dfrac{9}{7}

Since $\dfrac{a$1}{a2} = \dfrac{b1}{b2} \neq \dfrac{c1}{c2}, the two lines are parallel.

Hence, the two lines are parallel.

(iv) Given,

Equation 1: x - 2y = 0

a₁ = 1, b₁ = -2, c₁ = 0

Equation 2: 3x - 4y - 20 = 0

a₂ = 3, b₂ = -4, c₂ = -20

$\Rightarrow \dfrac{a$1}{a2} = \dfrac{1}{3}, \dfrac{b1}{b2} = \dfrac{-2}{-4} = \dfrac{1}{2}, \dfrac{c1}{c2} = 0

Since $\dfrac{a$1}{a2} \neq \dfrac{b1}{b2}, the two lines intersect at a point.

Hence, the two lines intersect at a point.

(v) Given,

Equation 1: 2x + 3y - 9 = 0

a₁ = 2, b₁ = 3, c₁ = -9

Equation 2: 4x + 6y - 18 = 0

a₂ = 4, b₂ = 6, c₂ = -18

$\Rightarrow \dfrac{a$1}{a2} = \dfrac{2}{4} = \dfrac{1}{2}, \dfrac{b1}{b2} = \dfrac{3}{6} = \dfrac{1}{2}, \dfrac{c1}{c2} = \dfrac{-9}{-18} = \dfrac{1}{2}

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

Hence, the two lines coincide.

(vi) Given,

Equation 1: x + 2y - 4 = 0

a₁ = 1, b₁ = 2, c₁ = -4

Equation 2: 2x + 4y - 12 = 0

a₂ = 2, b₂ = 4, c₂ = -12

$\Rightarrow \dfrac{a$1}{a2} = \dfrac{1}{2}, \dfrac{b1}{b2} = \dfrac{2}{4} = \dfrac{1}{2}, \dfrac{c1}{c2} = \dfrac{-4}{-12} = \dfrac{1}{3}

Since $\dfrac{a$1}{a2} = \dfrac{b1}{b2} \neq \dfrac{c1}{c2}, the two lines are parallel.

Hence, the two lines are parallel.