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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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If $ax+y-1=0$ ,$x+ay=0,x+y+a=0$ are concurrent then the value of $a^3-1$ is


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  • Three lines $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$ and $a_3x+b_3y+c_3=0$ are concurrent if $\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0$
Given that the lines $ax+y-1=0,\:\:x+ay=0\:\:and\:\:x+y+a=0$
are concurrent.
$\Rightarrow a^3-a+a-1=0$
$\Rightarrow a^3-1=0$
Hence (a) is the correct answer.
answered Feb 4, 2014 by sreemathi.v
edited Mar 19, 2014 by rvidyagovindarajan_1

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