### Chapter 3: Pair of Linear Equations in Two Variables

Q
##### Pair of linear equations in two variables CBSE NCERT Solutions

Question:

On comparing the ratios $\frac{a_1}{a_2},$   $\frac{b_1}{b_2}$  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 coincide:

(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

(i)  5x - 4y + 8 = 0; 7x + 6y - 9 = 0

We Compare equation 5x − 4y + 8 = 0 with $a_1x + b_1y + c_1 = 0$  and 7x + 6y – 9 = 0 with $a_2x + b_2y + c_2 = 0$.

We get  $a_1 = 5 , b_1 = - 4 , c_1 = 8$  and  $a_2 = 7, b_2 = 6 , c_2 = -9$

We have  $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$  , because  $\frac{5}{7} \neq \frac{-4}{6}$    .

Hence, lines have a unique solution which means they intersect at one point.

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

We compare equation 9x + 3y + 12 = 0  with $a_1x + b_1y + c_1 = 0$  and  18x + 6y + 24 = 0 with  $a_2x + b_2y + c_2 = 0$.

We get  $a_1 = 9 , b_1 = 3 , c_1 = 12$  and  $a_2 = 18, b_2 = 6 , c_2 = 24$

We have   $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$   , because   $\frac{9}{18} = \frac{3}{6} = \frac{12}{24} \Rightarrow \frac{1}{2} = \frac{1}{2} = \frac{1}{2}$  .

Hence, lines coincide.

(iii)  6x - 3y + 10 = 0; 2x - y + 9 = 0

Comparing equation 6x − 3y + 10 = 0 with $a_1x + b_1y + c_1 = 0$  and 2x – y + 9 = 0 with  $a_2x + b_2y + c_2 = 0$.

We get  $a_1 = 6 , b_1 = -3 , c_1 = 10$  and  $a_2 = 2, b_2 = -1 , c_2 = 9$.

We have   $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$   because of   $\frac{6}{2} = \frac{-3}{-1} \neq \frac{10}{9} \Rightarrow 3 = 3 \neq \frac{10}{9}$   .

Hence, lines are parallel to each other.

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