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 a unique solution is,

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

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

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Solution

The correct option is C $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
$T w o e q u a t i o n s w i l l h a v e a u n i q u e s o l u t i o n i f, a f t e r s o l v i n g t h e m, t h e s o l u t i o n i s v a l i d i . e t h e l i n e s a c t u a l l y i n t e r s e c t a t a p o i n t .$

$L e t t h e r e b e e q u a t i o n s a_{1} x + b_{1} y + c_{1} = 0.......... (i) a n d$
$a_{2} x + b_{2} y + c_{2} = 0......... (i i) .$

$M u l t i p l y i n g (i) b y a_{2} & (i i) b y a_{1} a n d e l i m i n a t i n g x b y s u b t r a c t i o n w e g e t,$

$y = \frac{a_{1} c_{2} - a_{2} c_{1}}{a_{2} b_{1} - a_{1} b_{2}} .$

$A g a i n M u l t i p l y i n g (i) b y b_{2} & (i i) b y b_{1} a n d e l i m i n a t i n g y b y s u b t r a c t i o n w e g e t$

$x = \frac{b_{2} c_{1} - b_{1} c_{2}}{a_{2} b_{1} - a_{1} b_{2}} .$

$I t i s c l e a r t h a t t h e s o l u t i o n (x, y) = (\frac{b_{1} c_{2} - b_{2} c_{1}}{a_{2} b_{1} - a_{1} b_{2}}, \frac{b_{2} c_{1} - b_{1} c_{2}}{a_{2} b_{1} - a_{1} b_{2}}) w i l l b e v a l i d i f a_{2} b_{1} - a_{1} b_{2} \neq 0 ⟹ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} .$

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

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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 a unique solution 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 a unique solution is,