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The condition for the pair of equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ to have no solution is

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a_{1} b_{2} - a_{2} b_{1} = 0

a_{1} b_{2} - a_{2} b_{1} \neq 0

a_{1} b_{1} - a_{2} b_{2} = 0

a_{1} b_{1} - a_{2} b_{2} \neq 0

answer is A.

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Detailed Solution

It is given the system of linear equation is,

a_{1} x + b_{1} y + c_{1} = 0

……(1)

a_{2} x + b_{2} y + c_{2} = 0

………(2)
Two equations will have no solution if in the graph they represent parallel lines.
Let there be equations,

a_{1} x + b_{1} y + c_{1} = 0 \dots \dots \dots . (i)

and

a_{2} x + b_{2} y + c_{2} = 0 . . . . . . . . . (ii) .

Multiplying

(i)

a_{2} (ii)

a_{1}

and eliminating

x

by subtracting we get,

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

Again multiplying

(i)

b_{2} (ii)

b_{1}

and eliminating by subtraction we get,

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

It is clear that the solution

(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}})

will be invalid if
The denominator is a 0 quantity.
i.e.

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

\Rightarrow a_{1} b_{2} - a_{2} b_{1} = 0

So the correct option is 1.

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