Question

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 coincident:
(i) \(5x – 4y + 8 = 0\) , \(7x + 6y – 9 = 0\) (ii) \(9x + 3y + 12 = 0\), \(18x + 6y + 24 = 0\) (iii) \(6x – 3y + 10 = 0\), \(2x – y + 9 = 0\)

Answer 1

(i) The given equations are \(5x − 4y + 8 = 0\) and \(7x + 6y − 9 = 0 \). Comparing these with \(a_1x +b_1y +c_1 =0 \) and \(a_2x +b_2y +c_2 =0\), we find: \(a_1=5 ,b_1=-4, c_1=8\) \(a_2=7 ,b_2=6, c_2 =-9\) The ratios are: \(\dfrac{a_1}{a_2} =\dfrac{ 5}{7}\) \(\dfrac{b_1}{b_2} = \dfrac{-4}{6} =\dfrac{-2}{3}\) Since \(\dfrac{a_1}{a_2} ≠ \dfrac{b_1}{b_2}\), the lines intersect at a single point, meaning there is a unique solution.

---

(ii) The given equations are \(9x + 3y + 12 = 0 \) and \(18x + 6y + 24 = 0 \). Comparing these with \(a_1x +b_1y +c_1 =0 \) and \(a_2x +b_2y +c_2 =0 \), we find: \(a_1=9 ,b_1=3 ,c_1=12\) \(a_2=18, b_2=6, c_2 =24\) The ratios are: \(\dfrac{a_1}{a_2} =\dfrac{9}{18}=\dfrac{1}{2}\) \(\dfrac{b_1}{b_2} =\dfrac{3}{6} =\dfrac{1}{2}\) \(\dfrac{c_1}{c_2} = \dfrac{12}{24} =\dfrac{1}{2}\) Since \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\), the lines are coincident, indicating infinite solutions.

---

(iii) The given equations are \(6x − 3y + 10 = 0\) and \(2x − y + 9 = 0 \). Comparing these with \(a_1x +b_1y +c_1 =0\) and \(a_2x +b_2y +c_2 =0\), we find: \(a_1=6, b_1=-3, c_1=10\) \(a_2=2, b_2=-1, c_2 =9\) The ratios are: \(\dfrac{a_1}{a_2} =\dfrac{6}{2} =\dfrac{3}{1}\) \(\dfrac{b_1}{b_2} = \dfrac{-3}{-1} = \dfrac{3}{1}\) \(\dfrac{c_1}{c_2} = \dfrac{10}{9}\) Since \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}≠\dfrac{c_1}{c_2}\), the lines are parallel and do not intersect, meaning there is no solution.