Do the following equations represent a pair of coincident lines? Justify your answer.
$ 3 x+\frac{1}{7} y=3 $$ 7 x+3 y=7 $

To find :

We have to find whether the given pairs of equations represent pairs of coincident lines.

Solution:

We know that,

The condition for coincident lines is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

(i) \(  3 x+\frac{1}{7} y=3 \)

$7(3x)+7(\frac{1}{7}y)=7(3)$

$21x+y-21=0$

\( 7 x+3 y-7=0 \)

Here,

$a_1=21, b_1=1, c_1=-21$

$a_2=7, b_2=3, c_2=-7$

Therefore,

$\frac{a_1}{a_2}=\frac{21}{7}=3$

$\frac{b_1}{b_2}=\frac{1}{3}$

$\frac{c_1}{c_2}=\frac{-21}{-7}=3$

Here,

$\frac{a_1}{a_2}≠\frac{b_1}{b_2}$

Hence, the given pair of linear equations has unique solution.

(ii) \( -2 x-3 y-1=0 \)

\( 6 y+4 x+2=0 \)

Here,

$a_1=-2, b_1=-3, c_1=-1$

$a_2=4, b_2=6, c_2=2$

Therefore,

$\frac{a_1}{a_2}=\frac{-2}{4}=\frac{-1}{2}$

$\frac{b_1}{b_2}=\frac{-3}{6}=\frac{-1}{2}$

$\frac{c_1}{c_2}=\frac{-1}{2}$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Hence, the given pair of linear equations represent coincident lines.  

(iii) \(  \frac{x}{2}+y+\frac{2}{5}=0 \)

$10(\frac{x}{2})+10(y)+10(\frac{2}{5})=0$

$5x+10y+4=0$

\( 4 x+8 y+\frac{5}{16}=0 \)

$16(4x)+16(8y)+16(\frac{5}{16})=0$

$64x+128y+5=0$

Here,

$a_1=5, b_1=10, c_1=4$

$a_2=64, b_2=128, c_2=5$

Therefore,

$\frac{a_1}{a_2}=\frac{5}{64}$

$\frac{b_1}{b_2}=\frac{10}{128}=\frac{5}{64}$

$\frac{c_1}{c_2}=\frac{4}{5}$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}$

Hence, the given pair of linear equations has no solution. 

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Updated on: 10-Oct-2022

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