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 a unique solution is -

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

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

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

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

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Solution

The correct option is B $a_{1} b_{2} - a_{2} b_{1} \neq 0$
For unique solution
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
$\Rightarrow a_{1} b_{2} - a_{2} b_{1} \neq 0$

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Similar questions

Q. The two lines a₁x + b₁y = c₁ and a₂x + b₂y = c₂ are perpendicular if
(a) a₁a₂ + b₁b₂ = 0
(b) a₁b₂ + a₂b₁
(c) a₁b₁ + a₂b₂ = 0
(d) a₁b₂ + a₂b₁ = 0

For the given pair of linear equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0,$ the solution $(x, y)$ is

For the pair of linear equations $a_{1} x$ + $b_{1} y$ + $c_{1} = 0$ and $a_{2} x$ + $b_{2} y$ + $c_{2} = 0$ , the solution set are $x = \frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ and $y = \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ . They can also be written as:

The solution (x,y) for the given pair of linear equations is $a_{1}$ x + $b_{1}$ y + $c_{1}$ =0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ =0

Q. Prove that the area of the parallelogram, the equations of whose
sides are

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

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

\frac{(d_{1} - c_{1}) (d_{2} - c_{2})}{a_{1} b_{2} - a_{2} b_{1}}

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The condition for the pair of equations a1x+b1y+c1=0 and a2x+b2y+c2=0 to have a unique solution is -

The correct option is B a1b2−a2b1≠0For unique solutiona1a2≠b1b2⇒a1b2−a2b1≠0

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 a unique solution is -

The correct option is B $a_{1} b_{2} - a_{2} b_{1} \neq 0$
For unique solution
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
$\Rightarrow a_{1} b_{2} - a_{2} b_{1} \neq 0$