A system of linear equations is given as follows :
$a_{1} x + b_{1} y + c_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0$
Condition for two lines to have infinitely many solutions is,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

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\frac{a_{1}}{a_{2}} \neq \frac{c_{1}}{c_{2}}

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\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

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None of these

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Solution

The correct option is B $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$L e t u s t a k e t w o e q u a t i o n s a s$
$a_{1} x + b_{1} y + c_{1} = 0........ (i) a n d a_{2} x + b_{2} y + c_{2} = 0....... (i i) .$

$I f t h e r e a r e i n f i n i t e m a n y s o l u t i o n s t h e l i n e s s h o u l d c o i n c i d e . F o r c o i n c i d e n c e o f t h e l i n e s t h e s l o p e s o f t h e t w o e q u a t i o n s s h o u l d b e e q u a l a n d t h e i n t e r c e p t s o f t h e t w o e q u a t i o n s s h o u l d b e e q u a l .$

$W r t i n g t h e e q u a t i o n s i n s l o p e - i n t e r c e p t f o r m w e h a v e,$

$(i) a s y = \frac{- a_{1}}{b_{1}} x + (\frac{- c_{1}}{b_{1}}) a n d (i i) a s y = \frac{- a_{2}}{b_{2}} x + (\frac{- c_{2}}{b_{2}}) .$

$N o w i f t h e l i n e s c o i n c i d e t h e n t h e s l o p e s a r e e q u a l . i . e \frac{- a_{1}}{b_{1}} = \frac{- a_{2}}{b_{2}} ⟹ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} . . . . . . . . (i i i) .$

$A l s o t h e i n t e r c e p t s a r e e q u a l . i . e (\frac{- c_{1}}{b_{1}}) = (\frac{- c_{2}}{b_{2}}) ⟹ \frac{c_{1}}{c_{2}} = \frac{b_{1}}{b_{2}} . . . . . . (i v) .$

$C o m p a r i n g (i i i) & (i v) w e h a v e \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} .$

$A n s - O p t i o n A .$

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A system of linear equations is given as follows :a1x+b1y+c1=0a2x+b2y+c2=0Condition for two lines to have infinitely many solutions is,

A system of linear equations is given as follows :
$a_{1} x + b_{1} y + c_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0$
Condition for two lines to have infinitely many solutions is,