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) \(5x − 4y + 8 = 0\)
      \(7x + 6y − 9 = 0 \)

Comparing these equations with \(a_1x +b_1y +c_1 =0 \)and \(a_2x +b_2y +c_2 =0\) , we obtain
\(a_1=5 ,b_1=-4, c_1=8\)
\(a_2=7 ,b_2=6, c_2 =-9\)

\(\dfrac{a_1}{a_2} =\dfrac{ 5}{7} \)
\(\dfrac{b_1}{b_2} = \dfrac{-4}{6} =\dfrac{-2}{3} \)

Since, \(\dfrac{a}{a_2}≠ \dfrac{b_1}{b_2}\) 

Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.

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(ii) \(9x + 3y + 12 = 0 \)
    \(18x + 6y + 24 = 0 \)

Comparing these equations with \(a_1x +b_1y +c_1 =0 \) and \(a_2x +b_2y +c_2 =0 \), we obtain
\(a_1=9 ,b_1=3 ,c_1=12\)
\(a_2=18, b_2=6, c_2 =24\)

\(\dfrac{a_1}{a_2} =\dfrac{9}{18}=\dfrac{1}{2}\)
\(\dfrac{b_1}{b_2} =\dfrac{3}{6} =\dfrac{1}{2}\)
\(\dfrac{c_1}{c_2} = \dfrac{12}{24} =\dfrac{1}{2}\)

Since \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\)

Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations.

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(iii)\(6x − 3y + 10 = 0\)
\(2x − y + 9 = 0 \)

Comparing these equations with \(a_1x +b_1y +c_1 =0\) and \(a_2x +b_2y +c_2 =0\) , we obtain
\(a_1=6, b_1=-3, c_1=10\)
\(a_2=2 b_2=-1, c_2 =9\)

\(\dfrac{a_1}{a_2} =\dfrac{6}{2} =\dfrac{3}{1}\)
\(\dfrac{b_1}{b_2} = \dfrac{-3}{-1} = \dfrac{3}{1} \)
\(\dfrac{c_1}{c_2} = \dfrac{10}{9}\)

Since,\( \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}≠\dfrac{c_1}{c_2}\)

Hence, the lines representing the given pair of equations are parallel to each other, and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.