Question

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

1. x and y are variables.
2. a, b and c are real numbers and
3. 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 possibilities can happen:

1. the two lines intersect at one point.
   
2. the two lines do not intersect (i.e. the lines are parallel to each other)
   
3. the two lines coincide (i.e. the lines have infinite number of solutions).
   

For two linear equations \[a_1x + b_1y + c_1 = 0\] and \[a_2x + b_2y + c_2 = 0\], \[\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\] ⇒ lines intersect each other ⇒ lines have a unique solution \[\frac{a_1}{a_2} = \frac{b_1}{b_2}\] ⇒ lines do not intersect at all ⇒ lines are parallel to each other ⇒ lines have no solution} \[\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\] ⇒ lines are coincident ⇒ lines have infinite solutions

On comparing the ratios \[\frac{a_1}{a_2}, \frac{b_1}{b_2}\text{ and }\frac{c_1}{c_2}\], find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.

1. 7x − 5y + 10 = 0
   6x + 2y − 15 = 0
2. 5x + 2y + 8
   15x + 6y + 24 = 0
3. 4x − 8y + 9 = 0
   2x − 4y + 7 = 0
4. x − 2y = 0
   3x − 4y − 20 = 0
5. 2x + 3y − 9 = 0
   4x + 6y − 18 = 0
6. x + 2y − 4 = 0
   2x + 4y − 12 = 0

Answer 1

1. Given the equations 7x − 5y + 10 = 0 and 6x + 2y − 15 = 0
   \[a_1 = 7, \ b_1 = -5, \ c_1 = 10\]
   \[a_2 = 6, \ b_2 = 2, \ c_2 = -15\]
   \[\frac{a_1}{a_2} = \frac{7}{6}, \quad \frac{b_1}{b_2} = \frac{-5}{2}\]
   Since \[\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\], the lines intersect at a point.
2. Given the equations 5x + 2y + 8 = 0 and 15x + 6y + 24 = 0:
   \[a_1 = 5, \ b_1 = 2, \ c_1 = 8\]
   \[a_2 = 15, \ b_2 = 6, \ c_2 = 24\]
   \[\frac{a_1}{a_2} = \frac{5}{15} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{8}{24} = \frac{1}{3}\]
   Since \[\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\], the lines are coincident.
3. Given the equations 4x − 8y + 9 = 0 and 2x − 4y + 7 = 0:
   \[a_1 = 4, \ b_1 = -8, \ c_1 = 9\]
   \[a_2 = 2, \ b_2 = -4, \ c_2 = 7\]
   \[\frac{a_1}{a_2} = \frac{4}{2} = 2, \quad \frac{b_1}{b_2} = \frac{-8}{-4} = 2, \quad \frac{c_1}{c_2} = \frac{9}{7}\]
   Since \[\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\], the lines are parallel.
4. Given the equations x − 2y = 0 and 3x − 4y − 20 = 0:
   \[a_1 = 1, \ b_1 = -2, \ c_1 = 0\]
   \[a_2 = 3, \ b_2 = -4, \ c_2 = -20\]
   \[\frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}\]
   Since \[\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\], the lines intersect at a point.
5. Given the equations 2x + 3y − 9 = 0 and 4x + 6y − 18 = 0:
   \[a_1 = 2, \ b_1 = 3, \ c_1 = -9\]
   \[a_2 = 4, \ b_2 = 6, \ c_2 = -18\]
   \[\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2}\]
   Since \[\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\], the lines are coincident.
6. Given the equations x + 2y − 4 = 0 and 2x + 4y − 12 = 0:
   \[a_1 = 1, \ b_1 = 2, \ c_1 = -4\]
   \[a_2 = 2, \ b_2 = 4, \ c_2 = -12\]
   \[\frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3}\]
   Since \[\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\], the lines are parallel.