On comparing the ratios \( \dfrac{a_1}{a_2} , \dfrac{b_1}{b_2} \) and \( \dfrac{c_1}{c_2} \), find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or are coincident.
(i) \(5x - 4y + 8 = 0\); \(7x + 6y - 9 = 0\)
(ii) \(9x + 3y + 12 = 0\); \(18x + 6y + 24 = 0\)
(iii) \(6x - 3y + 10 = 0\); \(2x - y + 9 = 0\)
For a pair of equations in two variables
\[ a_1x + b_1y + c_1 = 0, \quad a_2x + b_2y + c_2 = 0, \]
the nature of the pair of lines is decided as follows:
(i) \(5x - 4y + 8 = 0\) and \(7x + 6y - 9 = 0\)
Here
\(a_1 = 5, b_1 = -4, c_1 = 8;\) \(a_2 = 7, b_2 = 6, c_2 = -9.\)
Compute the ratios:
\[ \dfrac{a_1}{a_2} = \dfrac{5}{7}, \quad \dfrac{b_1}{b_2} = \dfrac{-4}{6} = -\dfrac{2}{3}, \quad \dfrac{c_1}{c_2} = \dfrac{8}{-9} = -\dfrac{8}{9}. \]
Clearly \(\dfrac{5}{7} \neq -\dfrac{2}{3}\). Hence
The two lines intersect at a point.
(ii) \(9x + 3y + 12 = 0\) and \(18x + 6y + 24 = 0\)
Here
\(a_1 = 9, b_1 = 3, c_1 = 12;\) \(a_2 = 18, b_2 = 6, c_2 = 24.\)
Ratios:
\[ \dfrac{a_1}{a_2} = \dfrac{9}{18} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{12}{24} = \dfrac{1}{2}. \]
All three ratios are equal, so
The two lines are coincident.
(iii) \(6x - 3y + 10 = 0\) and \(2x - y + 9 = 0\)
Here
\(a_1 = 6, b_1 = -3, c_1 = 10;\) \(a_2 = 2, b_2 = -1, c_2 = 9.\)
Ratios:
\[ \dfrac{a_1}{a_2} = \dfrac{6}{2} = 3, \quad \dfrac{b_1}{b_2} = \dfrac{-3}{-1} = 3, \quad \dfrac{c_1}{c_2} = \dfrac{10}{9}. \]
We have \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\) but this common value is not equal to \(\dfrac{c_1}{c_2}\). Therefore
The two lines are parallel.
Compare the ratios \(\tfrac{a_1}{a_2}\), \(\tfrac{b_1}{b_2}\), and \(\tfrac{c_1}{c_2}\). Unequal first two → intersect; first two equal but not the third → parallel; all equal → coincident. Apply this test to each pair above.