$\begin{array}{1 1}(a)\;x+4y=0,x+2y=0\\(b)\;x-4y=0,x-2y=0\\(c)\;x+2y=0,x-y=0\\(d)\;x+3y=0,2x+y=0\end{array}$

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Given equation is $x^2-6xy+8y^2=0$

Factorising this equation we get

$(x-4y)(x-2y)=0$

$\Rightarrow\:$ The two lines are given by $x-4y=0$ and $x-2y=0$

or

$x^2-6xy+8y^2=0$ can be factorized using

Dharacharya method as

$x=\large\frac{6y\pm \sqrt{(-6y)^2-4.8y^2}}{2}$

$x=3y\pm y\sqrt{9-8}$

$x=3y\pm y$

$x=4y$ or $x=2y$

$x-4y=0$ or $x-2y=0$

Hence (b) is the correct answer.

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