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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

$(a)\;0\qquad(b)\;3\qquad(c)\;5\qquad(d)\;6$

Toolbox:
• 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.
$\therefore\:\begin{vmatrix}a&1&-1\\1&a&0\\1&1&a\end{vmatrix}=0$
$\Rightarrow\:a(a^2-0)-1(a-0)-1(1-a)=0$
$\Rightarrow a^3-a+a-1=0$
$\Rightarrow a^3-1=0$
Hence (a) is the correct answer.
edited Mar 19, 2014